Question

what is the largest divisor that divides 47, 80 and 124, and leaves remainder 3.
(a) 12 (b) 11 (c) 10 (d) 3

Answer 1

Answer:

Required the largest divisor that divides 47, 80 and 124, and leaves remainder 3 is 11.

Step-by-step explanation:

Here given numbers are 47,80 and 124

We want to find largest divisor that divides 47, 80 and 124, and leaves remainder 3.

Required largest divisor that divides 47, 80 and 124, and leaves remainder 3 is LCM of (47-3),(80-3),(124-3)

Therefore HCF of 44,77,121

HCF means Highest Common Factor.

By prime factorization,

$44 = 2 \times 2 \times 11 \\ 77 = 7 \times 11 \\ 121 = 11 \times 11$

Here highest common factor of 44,77 and 121 is 11.

Therefore we can say 11 is the largest divisor that divides 47, 80 and 124, and leaves remainder 3.

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