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Question 62
Find the second smallest number which, when divided by 81 or 63, leaves a remainder of 7 in each case.
Solution
LCM of 81 and 63 = $$3\times3\times3\times3\times7$$
=Â 567
Here 567 is the smallest number which, when divided by 81 or 63, leaves a remainder of zero. For remainder '7' we need to added it.
But we need the second smallest number which, when divided by 81 or 63, leaves a remainder of 7 in each case. So the required number = $$567\times2+7$$
=Â $$567\times2+7$$
= 1134+7
=Â 1141
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