What is the greatest number that will divide 38, 45, 52 and leave as remainders 2, 3 and 4 respectively?

Correct option is A

Given:

We need to find the greatest number that will divide 38, 45, and 52 and leave remainders 2, 3, and 4 respectively.

Concept Used:

If a number N divides numbers A, B, and C leaving remainders R1, R2, and R3, then:

N is a divisor of (A−R1) ,(B−R2),(C−R3)

Solution:

Thus, we find HCF of (38−2),(45−3),(52−4)

38−2=36 ,45−3=42 ,52−4=48

HCF (36, 42, 48):

Prime factorization:

36 = $$2^2 \times 3^2$$​​

42 = $$2 \times 3 \times$$ 7

48 = $$2^4 \times 3$$​​

Common factors: 2 and 3

HCF = 2×3=6

Option (A) is right.