Find the smallest number which on being divided by 8, 12 and 15 leaves remainder 6, 10 and 13 respectively. But when divided by 23 no reminder is left
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Answer:
598
Step-by-step explanation:
Since (8-6) =(12-10) =(15-13) =2,
There fore, the required number =(LCM of 8,12,15)-2
= 120-2 =118
But by the given condition the number must be exactly divided by 23
The number =(multiple of 120)-2 = (120*5) - 2 =598 (since 598/23 = 6)
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