a and b are digits in a four-digit natural number 7a5b. If 7a5b is divisible by 18, how many different possible values can "a" have?
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If a number is divisible by 18, what two numbers must it be divisible by? A couple of possibiliies, but the obvious should be what? Then use divisibility rules to extrapolate the values. – Eleven-Eleven Jan 04 '14 at 13:27
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It should be divisible by 6 and 3. So we should find the possible values of b first then of a. Am I right? – Ally Jan 04 '14 at 13:29
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you could use 6 and 3, and you would have the same argument as below. – Eleven-Eleven Jan 04 '14 at 13:36
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Generally, coprime numbers must be used. For example, $24$ is divisible by $6$ and $3$, but it isn't divisible by their product, $18$. – JPLF Jan 04 '14 at 13:51
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It's okay. From my calculations a= 6,4,2,0,9,7 – Ally Jan 04 '14 at 13:59
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$7a5b = 7000+100a+50+b = 7050+100a+b$ – chharvey Jan 04 '14 at 14:16
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A number is divisible by $18$ iff it's divisible by $2$ and $9$. So, we must have $b \in \{0,2,4,6,8\}$ and $7+a+5+b$ divisible by $9$, since a number is divisible by $9$ iff the sum of its digits is divisible by $9$. I think you can solve it by now.
JPLF
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One way:
$$7a5b=7050+100a+b\equiv 12+10a+b\pmod{18}$$
Clearly, $b$ must be even
and $9$ must divide $12+10a+b=12+10a+b\iff 3+a+b$ must be divisible by $9$
For example,
if $b=0,3+a$ must be divisible by $9\implies a=6$ as $0\le a\le9\iff3\le a+3\le12$
if $b=2,5+a$ must be divisible by $9\implies a=4$
and so on
lab bhattacharjee
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Since it is divisible by $18$, $7a5b$ must be divisible by $9$ and $2$. Using divisibility rules, $$b \text{ must be even}$$ $$7+a+5+b \text{ must be divisible by 9}$$ So if $b$ is even it is of the form $2k$ for some integer $k$. So then $7+a+5+2k=9m$. Now use cases for $b$. For example, when $b=2$, $$7+a+5+2=14+a$$ So $a$ must be 4 since $14+4=18=9\cdot2$. Now, what about $a=4,6,8,10$? You can try them on your own.
Eleven-Eleven
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