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A number x is divisible by 7. When this number is divided by 8, 12 and 16. It leaves a remainder 3 in each case. The least value of x is:
- A148
- B149
- C150
- D147
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0
CCommunity
I have read it somewhere 147 is correct
0
CCommunity
LCM of 8, 12 and 16 = 48
Required number
= 48a + 3 which is divisible by 7.
x = 48a + 3
= (7 × 6a) + (6a + 3) which is
divisible by 7.
i.e. 6a + 3 is divisible by 7.
When a = 3, 6a + 3 = 18 + 3
= 21 which is divisible by 7.
x = 48 × 3 + 3 = 144 + 3 = 147
Required number
= 48a + 3 which is divisible by 7.
x = 48a + 3
= (7 × 6a) + (6a + 3) which is
divisible by 7.
i.e. 6a + 3 is divisible by 7.
When a = 3, 6a + 3 = 18 + 3
= 21 which is divisible by 7.
x = 48 × 3 + 3 = 144 + 3 = 147