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The reverse of the number 129 is 921, and these add to 1050, which is divisible by 30.
How many three-digit numbers have the property that, when added to their reverse,
the sum is divisible by 30?

Sagot :

Answer:

Step-by-step explanation:

Write the number as [tex]abc[/tex] and use the divisibility rules for 10 and 3. From the rule for 10, we know that the last digit of [tex]abc+cba[/tex] is [tex]o[/tex], hence [tex]a+c=10[/tex]. Thus the options are [tex]a=1[/tex] & [tex]c=9[/tex], [tex]a=2[/tex], & [tex]c=8[/tex], etc.

From the rule for 3, namely that the digit sum must be divisible by 3, we find that [tex]2a+2b+2c=20+2b[/tex]  is divisible by [tex]3[/tex]. By hand it's easy to check that [tex]b=2,5,8[/tex] are the only options that work.

So there are 9 choices for [tex]a[/tex] and [tex]c[/tex] and [tex]3[/tex] for [tex]b[/tex] , giving 27 total.

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