# Quant Boosters - Hemant Malhotra - Set 8

• Q26) The six digit number 24687X is divisible by 9 where X is a single digit whole number. Find X.

• Number is divisible by 9 so sum of digits should be divisible by 9
2+4+6+8+7+X =27+x should be divisible by 9
so x could bw 0 or 9 so we can't determine unique value here so Ans- can't be determined
question6-If a number 232x4y is divisble by 15 then which of the following value of x+y is not possible
approach = 232x4y is divisible by 15 so it will be div by 3 and 5
for 5- last digit should be o or 5 so y could be 0 or 5

case1-
when y=0
then 232x40 is div by 3 so sum of digit should be div by 3
2+3+3+x+4+0=12+x should be div by 3 so x could be, 0,3 ,6 and 9

case2- last digt is 5
232x45 div by 3
so 2+3+2+x+4+5 =16+x should be div by 3

so x could be 2,5,8

so these are the possible combination

• Q27) A number 12AB31 is divisible by 33, Which of the following cannot be the value of A+B ?
a) 11
b) 17
c) 5
d) 8
e) None of the above

• 12AB31 is div by 33
33 = 3 * 11 so number is divisible by 3 and 11
12AB31 is div by 3
so 1 + 2 + A + B + 3 + 1 div by 3
7 + A + B is div by 3
so A + B could be 2,5,8,11,14,17 (bcz A and B can't be greater than 9 so max sum=18)
now div by 11
so 1 - 2 + A - B + 3 - 1 shoud be div by 11
1 + A - B should be div by 11 so A - B = -1 or 10
but 10 is not possible bcz max value of A and B is 9 so difference couldn't be greater than 9
so A - B = -1 => B - A = 1
and A + B could be 2, 5, 8, 11, 14, 17
now use, B - A = 1 and A + B = 8
= > B=9/2 which is not possible so only odd values of A+B is possible so possible values are 5,11 and 17

• Q28) If 49x81y is completely divisible by 66 then how many pairs of (x, y) values are possible ?

• div by 66 so div by 2,3 and 11
div by 2 means last dgit y=0,2,4,6,8 possible
now div by 3 so 22+x+y div by 3
so 1+x+y div by 3 so possible values of x+y=2,5,8,11,14,17
now div by 11
4-9+x-8+1-y = -12+x-y div by 11
so x-y could be 12 or -10 or 1
but 12 and -10 not possible so x-y=1 so x=3 and y=2 possible value x=9 and y=8

• Q29) Let N = 111...111(73 times). When N is divided by 259, the remainder is a, and when N is divided by 32, the remainder is b. Then a+b is equal to
a. 6
b. 8
c. 253
d. None of these.

• 259 = 37 * 7
111111 ( 1 repeated 6 times ) is divisible by 259, 37 and 7.
now the given no. can be written as 111111000000... + 111111000000... + and so on..
but the last '1' is left out. because there are 73 ,ones in the question hence the remainder is 1.

as for 32. 32 = 2^5.
General rule is fro 2^n , check the last n digits for remainder. 7 is the rem.

So sum 7 + 1 = 8

• Q30) Consider a number formed by writing 2004 consecutive 9's (9999 … 2004 times). This number is not divisible by which of the following ?
a) 7
b) 13
c) 37
d) 101
e) None of these

• 999 = 37 * 27
9999 = 99 * 101
999999 = 999* 1001 = 999 * 11 * 91 = 999 * 11 * 13 * 7
2004 is a multiple of 3, 4 and 6.
The given number can be divided into groups of three 9's, four 9's or six 9's at a time.
The given number is divisible by 7, 13, 37 and 101.

61

61

61

61

64

61

61

71