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 = $$3\times3\times3\times3\times7$$

= 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 = $$567\times2+7$$

= $$567\times2+7$$

= 1134+7

= 1141

Share on Whatsapp Share on Whatsapp

Create a FREE account and get:

Free SSC Study Material - 18000 Questions
230+ SSC previous papers with solutions PDF
100+ SSC Online Tests for Free

SSC MTS 2022 Question Paper Shift 2 (July 15th)

Find the second smallest number which, when divided by 81 or 63, leaves a remainder of 7 in each case.

Free SSC Quant Questions

Login to your Cracku account

Hello! 👋

Ready to Unlock Your Success?

Please Enter Your OTP

Please enter the following details

Create a free Cracku account