The sum of the squares of the digits in a 3 digit number is 98.The sum of the first digit and the last digit is the middle digit. If the digits were reversed,the number is increased by 198.What is the number?
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Let the number be 100a+10b+c. We know a+c=b and a²+b²+c²=98 and 100c+10b+a-100a-10b-c=99(c-a)=198, meaning c-a=2. So we can write c=a+2, therefore a+a+2=b and b=2a+2.

Sum of squares: a²+4a²+8a+4+a²+4a+4=98 when we substitute for b and c.

Therefore 6a²+12a-90=0=a²+2a-15=(a-3)(a+5), and a=3. So b=2a+2=8 and c=a+2=5.

The number is 385.

CHECK

3²+8²+5²=9+64+25=98, 583-385=198.

 

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