Q1. Which of following numbers are perfect numbers?
1. 18 2. 36 3. 496
Ans. By Euclid’s method, If you add 1 + 2 + 4 + 8 + 16, the sum is 31, a prime number. Therefore 31 × 16 = 496 is another perfect number
You can also find the divisors and add them.
Q2. Resolve 2730 into prime factors.
Ans. Keep dividing by prime numbers, till remainder is prime,
2730 = 2 × 3 × 5 × 7 × 13
Q3. Resolve 32760 into prime factors.
Ans. Keep dividing by prime numbers, till remainder is prime,
32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13
Q4. Express 27/8 as a mixed fraction
Ans. Divide the numerator by denominator; note the multiplier, whatever remainder is left divide it with the original denominator. For 27/8, 24/8 = 3, and remainder left is 3, therefore 33/8 is the mixed fraction
Q5. Express 35 7/17as a improper fraction.
Ans. Here we need to multiply the denominator with the non-fraction part and add it to numerator and using same denominator.
For 35 7/17= (35 x 17 + 7)/17 = 602/17
Q6. Beginning with a single pair of rabbits, if every month each productive pair bears a new pair, which becomes productive when they are 1 month old, how many rabbits will there be after n months?
Ans. Imagine that there are xn pairs of rabbits after n months. The number of pairs in month n +1 will be xn (considering rabbits never die) plus the number of new pairs born. But new pairs are only born to pairs at least 1 month old, so there will be xn – 1 new pairs. Therefore answer is xn + 1 = xn + xn – 1.
Q7. Which of following numbers are divisible by 12.
(a) 188078 (b) 12496
(c) 3961815 (d) 13685
(e) 28008
Ans. Divisibility rule of 12, number has got to be divisible by 3 and 4
188078, sum of digits = 42, divisible by 3, last two digits not divisible by 4, rejected.
12496, sum of digits = 22, not divisible by 3, rejected
3961815, sum of digits = 33, divisible by 3, last two digits not divisible by 4, rejected.
13685, sum of digits = 23, not divisible by 3, rejected.
28008, sum of digits = 18, divisible by 3, last two digits divisible by 4, it is divisible by 12.
Q8. What maximum power of 6 would perfectly divide 50!?
Ans. It will be the number of 6’s in 50!, as 6 = 2 × 3, it will be equal to number of these pairs in 50!.
No. of 2’s (leave decimals)
= 50/2 + 50/22 + 50/23 + 50/24 + 50/25
onwards powers of 2 will exceed 50,
= 25 + 12 + 6 + 3 + 1 = 47
No. of 3’s(leave decimals)
= 50/3 + 50/32 + 50/33
onwards powers of 3 will exceed 50,
= 16 + 5 + 1 = 22
Since 3’s are 22, there will be 22 pairs, maximum power of 6 is 22.
Q9. Rationalize (√5 + √3)/(4 + √15).
Ans. Rationalizing with (4 + √15)
(√5 + √3)(4 - √15)/ (4 + √15)(4 - √15)
4√5 - 5√3 + 4√3 - 3√5)/(16 - 15 = (√5 - √3)
Q10. How many zeros will be there in the value of 25!?
Ans. The number of zeros are number’s of (5 × 2)’s
No. of 5’s(leave decimals)
= 25/5 + 25/52 onwards powers of 5 will exceed 25,
= 5 + 1 = 6, as twos are plenty, but 5’s are 6 therefore number of zeros are 6
Q11. What will be the unit’s digit in 12896?
Ans. 12896 = (12824) (12824) (12824) (12824)
Since we know multiple of 4 of power of 8, last digit is 6
Last digit = 6 × 6 × 6 × 6 = 6
Q12. How many different positive integers exist between 106 and 107, the sum of whose digits is equal to 2?
Ans. Between 10 and 100, that is 101 and 102, we have 2 numbers, 11 and 20.
Similarly, between 100 and 1000, that is 102 and 103, we have 3 numbers, 101, 110 and 200. Extrapolating the trend, between 106 and 107, one will have 7 integers whose sum will be equal to 2.
Q12. What least number must be added to 127561 so that it is exactly divisible by 28?
Ans. Least number to be added plus the remainder when divided by the given number should give the divisor. Here when we divide 127561 by 28, quotient is4555 and remainder is 21, so 21 + least number = 28, least number = 7
Q13. What is the remainder when 1044 × 1047 is divided by 33?
Ans. Here when 33 divide 1044, remainder is 21.
And when 33 divide 1047 is 24
Now the rule is the remainders multiplication has to be divided by divisor to get eh remainder
Therefore 21 × 24 = 504. Dividing by 33 remainder is 9, which is the answer.
Q14. What is the total number of different divisors including 1 and the number that can divide the number 6400?
Ans. The number of divisors (including 1 and itself) of a given number N where
N = Am * Bn * Co … where A,B,C are prime numbers are (1 + m) (1 + n) (1 + o)…
Now, 6400 = 26 * 4 * 25 = 28 * 52. Here m = 8 and n = 2
Therefore number of divisors are (8 + 1)(2 + 1) = 9*3 = 27.
Q15. How many strokes on computer keyboard are needed to type numbers from 1 to 999?
Ans. 1 to 9 needs 9 strokes
10 to 99 needs 180 (90 × 2) strokes
100 to 999 needs 2700 (900 × 3) strokes
Total = 2892 strokes
Q16. What is the reminder when 91 + 92 + 93 + ...... + 99 is divided by 6?
Ans. It is obvious that 6 is leaving remainder 3 with powers of 9, which is actually a property, but students can check by dividing first 2–3 numbers in the series, therefore total remainder is 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 27, when 27 is divided by 6, remainder is 3, which is the answer.
Q17. A certain number when successfully divided by 8 and 11 leaves remainders of 3 and 7 respectively. What will be remainder when the number is divided by the product of 8 and 11?
Ans. Solution to this question comes by a rule although there is a longish method using the choices also. Here divisor 1 = 8, divisor 2 = 11, remainder 1 = 3, remainder 2 = 7, by rule it is equal to divisor 1 × remainder 2 + remainder 1 = 8 × 7 + 3 = 59.
Q18. The smallest number which, when divided by 4,6 or 7 leaves a remainder of 2, is:
(a) 44 (b) 62
(c) 80 (d) 86 (CAT 1993)
Ans. (d) In the question it is asked, “The smallest number divided by 4, 6 and 7”, this clearly indicates that the number is the LCM of three numbers. The remainder of 2 would mean, the number is LCM + 2.
Therefore, required number = LCM of (4, 6, 7) + 2
LCM of 4,6,7, making prime factors, 2 × 2, 2 × 3, 7, is equal to 2 × 2 × 3 × 7 = 84 Required number = 84 + 2 = 86
Q19. If both 112 and 33 are factors of the number a* 4³ * 6² *1311, then what is the smallest possible value of a?
Ans. Here 62 = 32 22, therefore one power of 3 is short and 112 are not possible in given number, therefore a = 3 × 112
Q20. What is the last digit of 22^33^44^55^66^77
Ans. 22^33^44^55^66^77
It can be evaluated by just considering 2 instead of 22 and neglecting higher powers Any power of 33 × 4n = 3^4n ends in 1 ... that is ... it is of the form 5n + 1 thus 2^(5n + 1) as cyclicity of 2 is 5 .....we will get the last digit as 2 × 1 = 2
last digit of 66^77 = 6
last digit of 55^66^77 = 5
last digit of 44^55^66^77 = 44^(something)25 is same as 44^1 = 4
last digit of 33^44^55^66^77 = 1
last digit of 22^33^44^55^66^77 =2
Q21. The number of positive integers not greater than 100, which are not divisible by 2,3 or 5 is : (CAT 1993)
(a) 26 (b)18
(c) 31 (d) none of these
Ans. (a) There are 50 odd numbers less than 100 which are not divisible by 2.
Out of these 50 there are 17 numbers which are divisible by 3. [Total numbers divisible by 3 from 1 to 99 are 33, out of last number divisible is 99(33 × 3) is an odd number, so odd numbers divisible are 17] Out of remaining there are 7 numbers that are divisible by 5. [Total numbers divisible by 5 from 1 to 99 are 19, out of which 9 are even (10,20,...90), and three odd numbers are divisible by 3(15, 45, 75), therefore 7 are left]
Hence numbers which are not divisible by 2,3,5 = (50 – 17 – 7) = 26
Q22. What is the remainder when 2000^1000 is divided by 13?
Ans. 2000 could be split up as 2002 – 2, 2002 being a multiple of 13 so we worry but (–2)^1000 or (2)^1000 split that as 2^4 * (2^6)^6 so we have 16*(64)^6 in nr. i.e. 16*(65–1)^6
As 65 is divisible by 13, the remainder would be decided by 16*1/ 13 Þ 3 is the remainder
Q23. Let x < 0.50 , 0 < y <1 , z > 1. Given a set of numbers; the middle number, when they are arranged in ascending order, is called the median. So the median of number x , y and z would be: (CAT 1993)
(a) less than 1 (b) between 0 and 1
(c) greater than 1 (d) cannot say
Ans. (b) The median is the middle number in ascending order, given x < 0.50, 0 < y < 1, z > 1, the numbers could be arranged as (x, y, z) or (y, x, z). since x and y both are under 1, hence median will also lie between 0 and 1.
Q24. A number on being divided by 5 & 7 successively leaves the remainders 2 & 4 respectively. Find the remainder when the same number is divided by 35.
Ans. The number must be in the form of 5(7x + 4) + 2 = 35x +22 thus remainder = 22
Q25. What is the digit in the unit’s place of 251? (CAT 1998)
(a) 2 (b) 8
(c) 1 (d) 4
Ans. (b) The cycle of powers of two is 2,4,8,6 as last digit and repeat. As per that a power of 52(multiple of 4) has last digit of 6, therefore one behind 51 should have last digit of 8.
Q26. If a, a + 2 and a + 4 are prime numbers, then the number of possible solutions for a is: (CAT 2003)
(a) 1 (b) 2
(c) 3 (d) more than 3
Ans. (a) a, a + 2 , a + 4 are prime numbers. The number fits is 1, 3, 5 and 3, 5, 7 but post this nothing will fit. Now 1, 3, 5 are not prime numbers as 1 is not prime number.
So, only one possibility is there 3, 5, 7 for a = 3.
Q27. The product of all integers from 1 to 100 will have the following number of zeroes at the end: (CAT 1993)
(a) 20 (b) 24
(c) 19 (d) 22
Ans. (b) Number of zeros are the number of 5 × 2’s in the number, a multiplication of 1 to 100 is also 100!, here number of 5’s will be lesser than number of 2’s(obviously) and therefore total number of zeros will be the total number of 5’s. The number 5’s can be calculated as:
100/5 = 20
100/52 = 4, we will not divide by 53 = 125 as it is greater than 100
Total 5’s = 20+4 = 24, Thus, number of zeroes in the product of all the numbers from 1 to 100 is 24.
Q28. The product of 2 numbers is 7168...and the HCF is 16..find the numbers...
Ans. Divide 7168 with 16*16 we get quotient as 28 which can be represented as 1*28 or 4*7 (2*14 cannot be considered as it will give 32 hcf) Numbers are (16,448 or (64,112).
To solve these kind of problems, just follow these 3 steps
1. find the value of product/(HCF^2)
2. find the possible pairs of factors of value obtained in step 1
3. multiply the HCF with the pair of prime factors obtained in step 2
Q29. What is the smallest number which when increased by 5 is completely divisible by 8,11 and 24? (CAT 1994)
(a) 264 (b)259
(c) 269 (d) none of these
Ans. (b) The smallest number divisible is the least common factor, therefore LCM of three numbers 8, 11 and 24 = 2 × 2 × 2 × 3 × 11 = 264
Required number = LCM of ( 8,11,24 ) – 5
= (264 – 5 ) = 259.
Q30. If log7 log5 (vx + 5 + vx ) = 0, find the value of x. (CAT 1994)
(a) 1 (b) 0
(c) 2 (d) none of these
Ans. (b) log7log5 (vx +5 + vx ) = 0
log5 (vx + 5 + vx ) = (7)0 = 1 logax = y, then x = (a)y
= (vx +5 + vx) = (5)1 = 5
2vx = 0
thus x = 0
Q31. Let un + 1 = 2un + 1 , (n = 0, 1, 2,…) and u0 = 0. then u10 would be nearest to:
(a) 1023 (b) 2047
(c) 4095 (d) 8195 (CAT 1993)
Ans. (a) Un + 1 = 2Un + 1 (n = 0,1,2….)
Put n = 0, U1 = 1 n = 1 ,U2 = 3 n = 2, U3 = 7 n = 3, U4 = 15
The pattern is 1,3,7,15….., which can be written as 21 – 1, 22 – 1, 23 – 1, 24 -1, …..
Seeing the pattern it is clear that Un = 2n – 1
Hence U10 = (2)10 – 1 = 1023.
Q32. A,B and C are defined as follows:
A = ( 2.000004) ÷ [(2.000004)2 + ( 4.000008)],
B = ( 3.000003) ÷ [(3.000003)2 + ( 9.000009)],
C = ( 4.000002) ÷ [(4.000002)2 + ( 8.000004)],
Which of the following is true about the values of the above three expressions?
(a) all of them lie between 0.18 and 0.2
(b) A is twice of C
(c) C is the smallest
(d) B is the smallest (CAT 1997)
Ans. (d) Here A and C are in format of x/(x2 + 2x), as per this format the bigger the value of x, the lower is the value of expression. Therefore A > C. The first option is incorrect as per expression for x = 2, expression value is 0.25, The second option is incorrect clearly. Now for third and fourth option, B is in the format of x/(x2 + 3x), the value of x in C is greater than in B by almost 1. For all such values(with difference close to 1) B will be the smallest.
Q33. If n – 1 = x, where x is the product of four consecutive positive integers, then which of the following is/are true? (CAT 1999)
A. n is odd B. n is prime
C. n is a perfect square
(a) A and C only (b) A and B only
(c) A only (d) none of these
Ans. (a) Let’s do it for 1,2,3,4, which makes x = 1 x 2 × 3 × 4 = 24
Thus n = 1 + 24 = 25 ,we find that n is odd and a perfect square. This is true for any set of four consecutive positive integers.
Q34. The remainder , when ( 1523 + 2323 ) is divided by 19, is : (CAT 2004)
(a) 4 (b) 15
(c) 2 (d) 18
Ans. (c) an + bn is always divisible by a + b when n is odd.
Thus 1523 + 2323 is always divisible by 15 + 23 = 38. As 38 is a multiple of 19 , 1523 + 2323 is divisible by 19. So we get a remainder of 0.
Q35. A young girl counted in the following way on her fingers of her left hand. She started calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction calling the ring finger 6, middle finger 7, index finger 8, thumb 9 and then back to the, index finger for 10, middle finger for 11 and so on. She counted up to 1994. she ended on her: (CAT 1993)
(a) thumb (b) index finger
(c) middle finger (d) ring finger
Ans. (b) If the girl counts the way given in the question, then counting serial for the thumb will be 1,9,17,25……, a difference of 8, and trend of 1, 8x1 +1, 8x2 + 1, ….now if 1994 is divided by 8, it leaves remainder of 2, therefore it perfectly divides 1992. Since 1992 is multiple of 1993 will also fall on thumb. Hence number 1994 will end on her index finger.
Q36. Three consecutive positive even numbers are such that thrice the first number exceeds double the third by 2, the third number is : (CAT 1995)
(a) 10 (b) 14
(c) 16 (d) 12
Ans. (b) Let the three even numbers be x, (x + 2), (x + 4)
Then 3x – 2(x + 4) = 2, x = 10, numbers are 10, 12, 14
Thus the third number = 14.
Q37. Which is the least number that must be subtracted from 1856 so that the remainder when divided by 7,12,16 is 4? (CAT 1994)
(a) 137 (b) 1361
(c) 140 (d) 172
Ans. (d) Since the remainder is asked dividing by three numbers, we need to take LCM of three numbers and divide the number by it. LCM of (7,12,16) = 7 × 2 × 2 × 2 × 2 × 3 = 336. If we divide 1856 by 336 then remainder is 176(1856 × 336 × 5). It is given that remainder in this condition is 4. Hence the least number to be subtracted = (176 – 4) = 172.
Answer the questions based on following information (Q 38 to 40 )
A young girl roopa leaves home with x flowers, goes to the bank of a nearby river. On the bank of the river, there are four places of worship, standing in a row. She dips all the x flowers into the river, the number of flowers doubles. Then she enters the first place of worship, offers y flowers to the deity. She dips the remaining flowers into the river, and again the number of flowers doubles. She enters the second place of worship, offers y flowers to the deity. She dips the remaining flowers into the river, and again the number of flowers doubles. She enters the third place of worship, offers y flowers to the deity. She dips the remaining flowers into the river, and again the number of flowers doubles. Then she enters the fourth place of worship, offers y flowers to the deity. Now she is left with no flowers in hand.
(CAT 1999)
Q38. If roopa leaves home with 30 flowers , the number of flowers she offers to each deity is :
(a) 30 (b) 31
(c) 32 (d) 33
Questions (38 to 40 ) Let roopa had x flowers with her then
Balance before Flower offered Balance after
offering offering
Ist place 2x y 2x – y
IInd place 4x – 2y y 4x – 3y
IIIrd place 8x – 6y y 8x – 7y
IVth place 16x – 14y y 16x – 15y
16x – 15y = 0 so 16x = 15y y = 16x/15
Ans. (c) If x = 30, y = 16 × 2 = 32.
Q39. The minimum number of flowers that could be offered to each deity is:
(a) 0 (b) 15
(c) 16 (d) cannot be determined
Ans. (c) Minimum value for y is available for x = 15 y = 16/15 × 15 y = 16.
Q40. The minimum number of flowers with which roopa leaves home is:
(a) 16 (b) 15
(c) 0 (d) cannot be determined
Ans. (b) Minimum value of x is available for y = 16 thus x = 15/16. y = 15/16 × 16 = 15
Q41. 56 – 1 is divisible by : (CAT 1995)
(a) 13 (b) 31
(c) 5 (d) none of these
Ans. (b) Here (56 – 1) = (53)2 – (1)2 = (125)2 – (1)2
(125 + 1) (125 – 1 ) = 126 × 124, Here 124 is divisible by 31(31 × 4)
Therefore it is divisible by 31.
Q42. Suppose that you have a table with 8 rows and 8 columns and that the numbers from 1 to 64 are placed in the table in such a way that all the rows and columns add up to the same number. What is this number?
Ans42. The sum of 1 to 64 = 2080. Since there are 8 rows & the sum of each row is the same and the sum of all the rows is 2080, therefore the sum of each row is 2080/8 = 260
Q43. Ruby wanted to cut a cubical cake into 125 identical pieces applying minimum number of cuts. What is the number of cuts she has to apply?
Ans43. Since 125 = 5 × 5 × 5, the number of pieces required are 5,5,5 which will be created by cuts 4, 4, 4, therefore total number of cuts required are 12.
Q44. There are six locks exactly with one key for each lock. All the keys are mixed to each other. The maximum number of attempts needed to get the correct combination is
Ans44. Let there be six locks A, B, C, D, E and F. Then we need maximum 6 attempts to know the right key for A, and then 5 for B and then 4 for C and then 3 for D, and then 2 for E and none for F(since it is the only key left). Thus the total number attempts required = 6 + 5 + 4 + 3 + 2 = 20
Q45. In an arithmetic class, pupils are asked how many legs a hen, six dogs and seven palpigradis have in total. (“Palpigradi” is the Latin name of a certain animal). Alex says 46, Ben says 52, Cecilia says 66, Dora says 78, and Edith says 82. Who is right?
Ans45. If the number of legs of a palpigradi is x, then the total number of legs is 2 + 6.4 + 7.x = 7(x + 3) + 5. Thus the remainder of the number of legs divided by 7 is 5. 82 is the only number out of 46, 52, 66, 78 and 82 that leaves a remainder of 5 when divided by 7, therefore only Edith can be right. By solving the equation 2 + 6.4 + 7.x = 82, therefore a palpigradi has 8 legs.
Q46. A group of friends went on a holiday to a hill station. It rained for 13 days. But when it rained in the morning, the afternoon was lovely. And when it rained in the afternoon, the day was preceded by clear morning. Altogether there were11 nice mornings and 12 nice afternoons. How many days did their holiday last?
Ans46. Let the number of total mornings be M and the number of total afternoons be E. The wet mornings must be M – 11 and the wet afternoons are E – 12 So, M – 11 + E – 12 = 13 or M + E = 36, but M + E is one day so the number of days are 18.
Q47. A cuboid of dimensions 51, 85 and 102 cm is first painted by blue color then it is cut into minimum possible identical cubes. Find the total surface area of all those faces of cubes, which are not blue.
Ans47. As 51 = 17 × 3, 85 = 17 × 5, 102 = 17 × 6
Therefore the minimum possible number of cubes = 3 × 5 × 6 = 90.
The total surface area of non-blue faces = (Total surface area of all the cubes total surface area of cuboid) = 90 × (6 × 17 × 17) – 2 × (51 × 85 + 85 × 102 + 102 × 51) = 119646 cm2.
Q48. What is the least number of digits (including repetitions) needed to express 10^100 in decimal notation?
Ans48. By a certain rule 10n has n+1 digits, therefore the answer is 101,, you can validate the property by taking examples.
Q49. Consider any two-digit number whose digits are not zero and are not the same. What is the greatest integer that divides evenly the difference between the square of the number and the square of the reverse?
Ans49. The largest two digit number being 99 will not fit obviously as its reverse is the same, going backwards, you can easily find that 91 is the number where difference is the largest that is = 912 – 192 = 7920 Hence, the greatest integer is 7920.
Q50. Find all the natural numbers with the following property: If the last two digits of the square of the number are interchanged, the result is the square of the next natural number.
Ans50. As per the question, if the last two digits of n2 are A and B (in this order) then the last two digits of (n+ 1)2 are B followed by A. Therefore 2n + 1 = (n + 1)2 – n2 = (10B + A) – (10A + B) =9 (B – A).
It immediately follows that B–A is odd. A = 0 is impossible, as in a square number ending in 0 the last but one digit is also 0, which makes B = 0, but 2n + 1 cannot be 0(being odd). Therefore B > A. As B – A is odd, B –A{1, 3, 5, 7}. Hence n{4, 13, 22, 31}. The squares of the corresponding pairs are 16 and 25(for 4 and 5); 169 and 196(for 13 and 14); 484 and 529(for 22 and 23); 961 and 1024(for 31 and 32). The requirement is only satisfied by one of them: 13.
Q51. The benches of the Hall of the Parliament of India are arranged in a rectangle of 10 rows of 10 seats each. All the 100 MPs have different salaries. Each of them asks all his neighbours (sitting next to, in front of, or behind him, as well as those four seated diagonally in front of or behind him, i.e. 8 members at most) how much they earn. They feel a lot of envy towards each other: an MP is content with his salary only if he has at most one neighbor who earns more than himself. What is the maximum possible number of MPs who are satisfied with their salaries?
Ans51. Represent the assembly hall of the parliament by a 10 by 10 table. It is easy to see that at most two out of the four MP’s seated in any 2 by 2 square may be satisfied with their salaries. Since the 10 by 10 table can be divided into 25 2 × 2 squares, in each of which there may be at most two satisfied MP’s, there may be at most 50 such members of the whole parliament.
If the MP’s seated in the even-numbered rows all have lower salaries than those in the odd-numbered rows, and within each odd-numbered row the salaries decrease left to right, then there will be exactly 50 satisfied MP’s.
Q52. Two players play the following game. They take turns in taking matches from a heap that initially contains 7 matches. In each step, a player can take one, two or three matches. The game continues until there are no matches left. The winner is the player holding an even number of matches at the end. Which player has a winning strategy, the one who starts or his opponent? How should he play in order to win?
Ans52. As 7 is an odd number, only one player can win. The opening player has a winning strategy. Suppose he opens by taking 3 matches. If his opponent takes 1 match in the next step, he will win by taking the remaining 3. If his opponent removes 2 matches, he can take 1 of the remaining 2. Finally, if the opponent takes 3 matches, he has the remaining 1, and he wins with 4 matches.
Q53. In the sequence obtained by omitting the squares from the sequence of natural numbers, what is the 2001st term?
Ans53. As 442 = 1936 < 2001 < 2025 = 452, there are 44 square numbers up to 2001. Hence 2001 is the 2001 – 44 = 1957th member of the sequence. The first square number following 2001 is 2025 = 452, then 462 = 2116. Thus 2001 is followed by 23 non-square numbers in a row, the largest of which is 2024, in the 1957 + 23 = 1980th place. Then 2115–2025 = 90 non-square numbers follow: 2026, 2027, ...; the 21th of which, 2026 + 20 = 2046 is the 2001st member of the sequence.
Q54. On what condition will the product of a number ending in 9 and a number ending in 7 end in 63?
Ans54. Let the numbers in question be A=10a+9, B = 10b + 7 (a and b are the digits preceding the last digit). Then AB = 20(7a + 9b) + 63 ends in 63 if and only if 7a + 9b = 10b + (7a – b) is divisible by 10, that is, if b equals the last digit of 7a. Thus the last two digits of A and B is given by one of the following number pairs: (09;07), (19;77), (29;47), (39;17), (49;87), (59;57), (69;27), (79;97), (89;67), (99;37).
Q55. Is there a square number in which the last two digits are both odd?
Ans55. The number in question can only be the square of an odd number, of the form (2k + 1)2 = 4k(k + 1) + 1, thus it leaves a remainder of 1 when divided by 4 (or even by 8). The last digit of an odd square number may be 1, 5 or 9, which now means that the remainder is 11, 15 or 19 when the number is divided by 20. Such numbers, however leave a remainder of 3 when divided by 4, and that contradicts to the observation above. Therefore, the last two digits of a square number cannot both be odd.
Q56. Anna was bored in class, and to kill time she made a list of integers. Starting with a certain number, she obtained the next number by either adding or multiplying the digits of the previous number on the list. She continued writing down numbers henceobtained, and observed that all the numbers were odd. With how many initial values at most six digits is that possible?
Ans56. Let us call the numbers of the required property “oddy”. In a “oddy” number, every digit is odd, since the product of the digits is odd. Thus there are 5 one-digit “oddy” numbers. There are none with two digits, because the sum of two odd digits is always even. For the same reason, there are none with four or six digits either. For three-digit “oddy” numbers: Note that both the sum and the product of the three odd digits must be “oddy”, which cannot be a two-digit number, therefore the sum of the digits (as well as their product) must be at most 9. Thus the smallest is 1, and the three digits must me 3 ones, 2 ones and any one of 3, 5 and 7, or 1 one with 2 threes. Hence there are 1 + 9 + 3 = 13 three-digit “oddy” numbers. Five-digit “oddy” numbers can be counted similarly; the sum of their digits is at most 45, and being “oddy”, the product is at most 9, the possible sets of digits are 5 ones; 4 ones and 1 three; 4 ones and 1 five ; 3 ones and 2 threes. That makes 1 + 5 + 5 + 10 = 21 five-digit “oddy” numbers altogether, and thus the total number of “oddy” numbers of at most six digits is 5 + 13 + 21 = 39.
Q57. We decreased a six-digit number by the sum of its digits, and went on repeating same procedure with the resulting number. Is it possible that we obtained 2002 in this way?
Ans57. The remainder of a number divided by nine equals the remainder of the sum of its digits. Thus the difference of the number and the sum of the digits is divisible by nine. Since 2002 is not divisible by nine, it cannot be obtained in the way described.
Q58. 100 cards are numbered 1 to 100 (each card different) and placed in 3 boxes (at least one card in each box). How many ways can this be done so that if two boxes are selected and a card is taken from each, then the knowledge of their sum alone is always sufficient to identify the third box?
Ans58. With a suitable labeling of the boxes as A, B, C, there are 4 cases to consider:
Case 1. A contains 1; B contains 2; C contains 3
Case 2. A contains 1, 2
Case 3. A contains 1, 3; B contains 2
Case 4. A contains 1; B contains 2, 3.
We show that Cases 1 and 4 each yield just one possible arrangement and Cases 2 and 3 none.
In Case 1, it is an easy induction that n must be placed in the same box as its residue (in other words numbers with residue 1 mod 3 go into A, numbers with residue 2 go into B, and numbers with residue 0 go into C). For (n + 1) + (n – 2) = n + (n – 1). Hence n + 1 must go in the same box as n – 2 (if they were in different boxes, then we would have two pairs from different pairs of boxes with the same sum). It is also clear that this is a possible arrangement. Given the sum of two numbers from different boxes, take its residue mod 3. A residue of 0 indicates that the third (unused) box was C, a residue of 1 indicates that the third box was A, and a residue of 2 indicates that the third box was B. Note that this unique arrangement gives 6 ways for the question, because there are 6 ways of arranging 1, 2 and 3 in the given boxes.
In Case 2, let n be the smallest number not in box A. Suppose it is in box B. Let m be the smallest number in the third box, C. m – 1 cannot be in C. If it is in A, then m + (n – 1) = (m – 1) + n. Contradiction (m is in C, n – 1 is in A, so that pair identifies B as the third box, but m – 1 is in A and n is in B, identifying C). So m – 1 must be in B. But (m – 1) + 2 = m + 1. Contradiction. So Case 2 is not possible.
In Case 3, let n be the smallest number in box C, so n – 1 must be in A or B. If n – 1 is in A, then (n – 1) + 2 = n + 2. Contradiction (a sum of numbers in A and B equals a sum from C and A). If n – 1 is in B, then (n – 1) + 3 = n + 2. Contradiction ( a sum from B and A equals a sum from C and B). So Case 3 is not possible.
In Case 4, let n be the smallest number in box C. n – 1 cannot be in A, or (n – 1) + 2 = 3 + n (pair from A, B with same sum as pair from B, C), so n – 1 must be in B. Now n + 1 cannot be in A (or (n + 1) + 2 = 3 + n), or in B or C (or 1 + (n + 1) = 2 + n). So n + 1 cannot exist and hence n = 100. It is now an easy induction that all of 4, 5, ... 98 must be in B. For given that m is in B, if m + 1 were in A, we would have 100 + m = 99 + (m + 1). But this arrangement (1 in A, 2 – 99 in B, 100 in C) is certainly possible: sums 3 /0 100 identify C as the third box, sum 101 identifies B as the third box, and sums 102-199 identify A as the third box. Finally, as in Case 1, this unique arrangement corresponds to 6 ways of arranging the cards in the given boxes.
Q59. Given an n × n square board, with n even. Two distinct squares of the board are said to be adjacent if they share a common side, but a square is not adjacent to itself. Find the minimum number of squares that can be marked so that every square (marked or not) is adjacent to at least one marked square.
Ans59. Let n = 2m. Color alternate squares black and white (like a chess board). It is sufficient to show that m(m + 1)/2 white squares are necessary and sufficient to deal with all the black squares. This is almost obvious if we look at the diagonals. Look first at the odd-length white diagonals. In every other such diagonal, mark alternate squares (starting from the border each time, so that r + 1 squares are marked in a diagonal length 2r + 1). Now each black diagonal is adjacent to a picked white diagonal and hence each black square on it is adjacent to a marked white square. In all 1 + 3 + 5 + ... + m–1 + m + m – 2 + ... + 4 + 2 = 1 + 2 + 3 + ... + m = m(m + 1)/2 white squares are marked. This proves sufficiency. For necessity consider the alternate odd-length black diagonals. Rearranging, these have lengths 1, 3, 5, ... , 2m – 1. A white square is only adjacent to squares in one of these alternate diagonals and is adjacent to at most 2 squares in it. So we need at least 1 + 2 + 3 + ... + m = m(m + 1)/2 white squares. Which is n/2 (n/2 + 1).
Q60. Find all pairs (a, b) of positive integers that satisfy ab2 = ba.
Ans60. Notice first that if we have am = bn, then we must have a = ce, b = cf, for some c, where m= fd, n= ed and d is the greatest common divisor of m and n. [Proof: express a and b as products of primes in the usual way.]
In this case let d be the greatest common divisor of a and b2, and put a = de, b2 = df. Then for some c, a = ce, b = cf. Hence f ce = e c2f. We cannot have e = 2f, for then the c’s cancel to give e = f. Contradiction. Suppose 2f > e, then f = e c2f–e. Hence e = 1 and f = c2f-1. If c = 1, then f = 1 and we have the solution a = b = 1. If c = 2, then c2f–1 = 2f > f, so there are no solutions.
Finally, suppose 2f < e. Then e = f ce-2f. Hence f = 1 and e = ce–2. ce–2 = 2e–2 = e for e = 5, so we must have e = 3 or 4 (e > 2f = 2). e = 3 gives the solution a = 27, b = 3. e = 4 gives the solution a = 16, b = 2.
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