ONLY AN ODD NUMBER WILL LEAVE REMAINDER 1 WHEN DIVIDED BY 4, SO WE ACTUALLY WANT ODD NUMBERS WHICH WILL LEAVE REMAINDER 4 WHEN DIVIDED BY 7 AND REMAINDER 1 WHEN DIVIDED BY 4.
25,53,81,109

ANS:B

Let the number be 7a+4 = 4b+1 ( When p is divided by 7, the remainder is 4. When p is divided by 4, the remainder is 1)
Equating we get, 4b-7a = 3
Solving for integral value of a, b ; we get min value of (a, b) = (3, 6)
hence number is 7*3+4 = 25
the smallest number is 25
for other numbers add (LCM of 7 & 4)28 to 25 to get other numbers. hence others nos are 25, 53,81,109,137 and so on...
Numbers smaller than 120 will be 25, 53,81,109.
hence there will be 4 such numbers.

We have the following two equations:

p = 7Q + 4 where Q is a whole number.

So, p can be values such as 4, 11, 18, 25, …, 7(16) + 4 = 116.

and

p = 4S + 1 where S is a whole number.

So, p can be values such as 1, 5, 13, 14, 21, 25, …, 4(29) + 1 = 117.

We see that 25 is the first matching value of p. Since the LCM of 7 and 4 is 28, the next will be 25 + 28 = 53, and the next will be 53 + 28 = 81, and the final matching value will be 81 + 28 = 109, so there are a total of 4 values of p less than 120 that satisfy the two criteria.

Alternate Solution:

We have

p = 7Q + 4

for some Q and

p = 4S + 1

for some S.

Then, p + 3 = 7Q + 7 = 4S + 4 is divisible by both 7 and 4. Let’s list the numbers less than 120 that are divisible by 7 and 4, i.e., by 28: 0, 28, 56, 84, 112. However, these are values for p + 3; thus, the corresponding values for p are -3, 25, 53, 81, and 109. p can take any value in this list except for -3; therefore, there are four values in total.

Answer: B
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