Find the smallest no which leaves remainders 8 and 12 when divided by 28 and 32 respectively
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Here, 28-8=20 and 32-12=20.
∴, the required number will be 20 less than the LCM of 28 and 32.
28=2×2×7
32=2×2×2×2×2
∴, LCM =2×2×7×2×2×2=224
∴, the required number is =224-20=204
∴, the required number will be 20 less than the LCM of 28 and 32.
28=2×2×7
32=2×2×2×2×2
∴, LCM =2×2×7×2×2×2=224
∴, the required number is =224-20=204
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