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The least number which when divided by 16, 18, 20 and 25 leaves 4 as remainder in each case but when divided by 7 leaves no remainder is
LCM of 16, 18, 20 and 25 = 3600∴ Required number = 3600K + 4 which is exactly divisible by 7 for certain value of K.When K = 5,
Required number = 3600 × 5 + 4 = 18004 , which is exactly divisible by 7.
Find the least number which when divided by 16,18,20 and 25 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.
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The smallest number which when divided by 20,25,35 and 40 leaves a remainder of 14,19,29 and 34 respectively, is 1394 . How?
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Now LCM of 20, 25, 35 and 40 -
20 = 2^2 * 5
25 = 5^2
35 = 5*7
40 = 2^3 * 5
therefore LCM = 2^3 * 5^2 * 7 = 1400
As i wrote above to get the final answer we have to subtract common value from LCM.
Final answer = 1400 - 6 = 1394
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