121. ( 6+ 5+ 3 + 8+) – (X + 2 6) = (14 – X). Now, (14 –X ) is divisible by 11. When X = 3.
122. (4 + 6 + 1 +X +y + 0 ) = [17 + ( X + y )]. Also, ( 0 + X + 7 ) –( y + 4 ) = ( X –y – 3).
Now, [17+ ( x + y )] must be divisible by 3 and ( x – y -3 ) is either 0 or divisible by 11.
Clearly, X =8 and = 5 satisfy both the conditions.
123. (a) 639 is not divisible by 7. (b) 2079 is divisible by 3, 7, 9 and 11.
(c) 3791is not divisible by 3. (d) 37911 is not divisible by 9.
Correct answer is (B).
124. Since 4864 is divisible boy4, so 9P2 must be divisible by 3.
(11 + P) must be divisible by3.
Least value of P is 1.
125. The required number should be divisible by 3 and 8.
(a) 718 is not divisible by8. (b) 810 is not divisible by 8.
(C) 804 is not divisible by 8. (d). sum of digits = 27, which is divisible by 3.
And, 736 is divisible by 8. So, given number is divisible by 3 and 8.
126. The given number should be divisible boy both 9 and 8.
(4 + 2 + 5 + 7 + 3 + X) = (21+ X) is divisible boy 9 and (73 X) is divisible by 9.
Also, (y + 3 + 2 + 2 + 3 ) – ( X + 1 + 2 + 4) = ( y – X + 3 ) must be 0 or divisible by 11.
X + y = 10 and y- X + 3= 0.
Clearly, X = 1 =9 satisfy both these equations.
127. Since 638 Xy is divisible by 5 as well as 2, so y = 0.
Now, 653 X0 must be divisible bioy8.
So, 3X0 must be divisible by 8. This happens when X=2.
X + y =( 2 + 0) = 2.
128. A number is divisible by 132, if it is divisible by each one of 11, 3 and 4. Clearly, 968 are not divisible by 3. None of 462 and 2178 is divisible by 4.
Also, 5148 is not divisible by 11..
Each one of remaining 4 is divisible by each one of 11, 3 and 4 and therefore, by 132.
129. Clearly, 6897 is divisible by both 11 and 19.
130. None of the numbers in (a) and (c) is divisible by2.
Number in (b) is to divisible by 3.
Clearly, 510510 is divisible by each prime number between by 1 and 17.
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