find the smallest no which leaves remainder 8 and 12 when divided by 28 and 32 respectively. Answer fast. And follow me l will follow back you.
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Given that the smallest no. When divided by 28 & 32 leaves remainder 8 & 12. Therefore, 28–8=20 & 32–12=20 are divisible by the required number will be 20 less than the LCM of 28 and 32. Hence, the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 is 204…
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