What is the largest number which when divides 64, 78 and 115 leaves 4, 6 and 7 as remainders, respectively?
Correct Answer: Option D
Let the number to be found is x.
When 64 is divided by x, remainder is 4
⇒ (64 – 4 = 60) is completely divisible by x.
Similarly, 78 is divided by x, the remainder is 6
⇒ (78 – 6 = 72) is completely divisible by x.
Also, when 115 is divided by x, remainder is 7
⇒ (115 – 7 = 108) is completely divisible by x.
∵x is the largest number that divides 60, 72 and 108, x must be the HCF of 60, 72 and 108.
60 = 2 × 2 × 3 × 5
72 = 2 × 2 × 2 × 3 × 3
108 = 2 × 2 × 3 × 3 × 3
∴ HCF = 2 × 2 × 3= 12
∴ required number = 12
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