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# Suppose that the least common multiple of the first 25 positive integers is equal to 26A7114B4C0. Find 100*A + 10*B + C.

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Suppose that the least common multiple of the first 25 positive integers is equal to 26A7114B4C0. Find 100*A + 10*B + C.

Jul 25, 2018

### 2+0 Answers

#1
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Find 100*A + 10*B + C.

[100*7] + [10*4 + 0] =

700        +    40            = 740

Jul 26, 2018
#2
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Suppose that the least common multiple of the first 25 positive integers is equal to

26A7114B4C0.

Find 100*A + 10*B + C.

$$\begin{array}{|rcll|} \hline && \text{lcm}(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25) \\ &=& 26{\color{red}7}7114{\color{red}4}4{\color{red}0}0 \\ && 26{\color{red}A}7114{\color{red}B}4{\color{red}C}0 \\\\ && A=7 \\ && B=4 \\ && C= 0\\ && 100*A + 10*B + C \\ &=& 100*7 + 10*4 + 0 \\ &=& 700 + 40 + 0 \\ &\mathbf{=}& \mathbf{740} \\ \hline \end{array}$$ Jul 26, 2018