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# help

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Find the smallest positive multiple of 7 that leaves a remainder of 4 when divided by 6, 9, 15, and 18.

Dec 2, 2019

#1
+1

112 mod [6, 9, 15, 18] = 4

Dec 2, 2019
#2
+23910
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Find the smallest positive multiple of 7 that leaves a remainder of 4 when divided by 6, 9, 15, and 18.

$$\begin{array}{|rcll|} \hline 7x &\equiv& 4 \pmod{6} \\ 7x &\equiv& 4 \pmod{9} \\ 7x &\equiv& 4 \pmod{15} \\ 7x &\equiv& 4 \pmod{18} \\ \hline 7x &\equiv& 4 \pmod{ lcm(6,9,15,18) } \quad | \quad lcm(6,9,15,18) = 90 \\ 7x &\equiv& 4 \pmod{ 90 }\quad | \quad 4\equiv 94\equiv 184\equiv 274\equiv \underbrace{364}_{ \text{divisible by 7} }\pmod{ 90 } \\ 7x &\equiv& 364 \pmod{ 90 } \quad | \quad : 7 \\ \mathbf{ x } &\equiv& \mathbf{ 52 \pmod{ 90 } } \\\\ x &=& 52+90z \qquad z\in \mathbb{Z} \\ x_{\text{min}} &=& 52 + 90* 0 \\ x_{\text{min}} &=& 52 \\ \hline \end{array}$$

The smallest positive multiple of 7 is $$52* 7 = \mathbf{364}$$

Thank you Guest!

Dec 3, 2019
edited by heureka  Dec 4, 2019
edited by heureka  Dec 4, 2019
edited by heureka  Dec 4, 2019
#3
+1

Both of the above answers are wrong!

112 mod 15 = 7 NOT 4

The smallest positive multiple of 7 that satisfies the 4 congruences is 364:

The LCM of 90m + 4, where m=0, 1, 2, 3.....etc.

Since the smallest multiples of 7 that satisfies the 4 congruences is:

4 x 90 + 4 = 364.

364 mod 6 = 4

364 mod 9 = 4

364 mod 15 = 4

364 mod 18 = 4

Dec 3, 2019
edited by Guest  Dec 4, 2019