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I would like it if someone could give me pointers on solving problems like these. And why was 4 the answer here ?

If $a=4b+26$ and $b$ is positive , then a could be divisible by all of following except

a)2 b)4 c)5 d)6 e)7

Edit:. I know by taking b=4 its divisible by 7 . However is there any other way by which we could tell its divisible by 7 other than randomly plugging in a no and testing ? Also I am using the current method to check if the expression is divisible by other options , is this method correct

Check by 2: 4b mode 2 and 26 mod 2 = 0 hence divisible by 2

For numbers greater than 4 the expression is doubled to $a=8b + 52$

Check by 5: 8b mode 5 = 3 while 52+3 mod 5 = 0 hence divisible by 5

Check by 6: 8b mode 6 = 2 while 52+2 mod 6 = 0 hence divisible by 6

Check by 4: 4b mode 2 =0 while 26 mod 2 $\not=$ 0 hence not divisible by 4

I need to know if this checking mechanism is correct ? is so why doesn't it work for 7 ? How could I check for 7 without plugging in and testing values ?

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    $0=4b+26=4b+5 \pmod 7 \implies 4b=2 \pmod 7$. Since $\gcd(4,7)=1$, there is a solution.(e.g. b=4)2017-12-30

3 Answers 3

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Suppose $a$ is divisible by $4$, so $a=4c$ for some integer $c$. Then $26=a-4b=4(c-b)$ is divisible by $4$, which is clearly not the case. Thus we have a contradiction.

In general, you should approach these problems using modular arithmetic. In this case we have $\begin{align} a &\equiv 4b+26 \mod 4\\ &\equiv 0b+2\mod 4\\ &\equiv 2\mod 4 \end{align}$ and so clearly $a$ is not divisible by $4$, as then we'd have $0\equiv 2\bmod 4$.

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It could never be divisible by 4, since $4b \mod 4 = 0$ for any integer $b$, and $26 \mod 4 = 2$.

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Hint $\rm\,\ d\:|\:4b\!+\!26\:\Rightarrow\: n\,d-4\,b = 26\:\Rightarrow\: gcd(d,4)\:|\:26\iff 4\nmid d$

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    Thanks for clearing that up.2012-07-23