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A number has remainder 13 when divided by 16, and has remainder 122 when divided by 125. What is the minimum four digit number?

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N=16k+13=125K1+122 16k=125k1+109 k=(125k1+109)/16 k=(112k1+96+13k1+13)/16, which implies k1 is 15 Therefore, the number is 125*15+122=1997 Alternatively, (16-13)=(125-122)=3, common difference of 3. Take lcm of 16 and 125 and subtract 3 to get the answer. As 16 and 125 are co prime numbers, lcm of 16...
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N=16k+13=125K1+122 16k=125k1+109 k=(125k1+109)/16 k=(112k1+96+13k1+13)/16, which implies k1 is 15 Therefore, the number is 125*15+122=1997 Alternatively, (16-13)=(125-122)=3, common difference of 3. Take lcm of 16 and 125 and subtract 3 to get the answer. As 16 and 125 are co prime numbers, lcm of 16 and 125 is 16*125=2000; 2000-3=1997. read less
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(16 x 125) - 3 = 1997
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(16*125) - 3
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