Three-digit 5312

Find the smallest four-digit number abcd such that the difference (ab) 2− (cd) 2 is a three-digit number written in three identical digits.

Correct answer:

x =  2017

Step-by-step explanation:

x1=9491 r1=942912=555 x2=7771 r2=772712=888 x3=6051 r3=602512=999 x4=5952 r4=592522=777 x5=5853 r5=582532=555 x6=5754 r6=572542=333 x7=5655 r7=562552=111 x8=4331 r8=432312=888 x9=4034 r9=402342=444 x10=2611 r10=262112=555 x11=2314 r11=232142=333 x12=2017 r12=202172=111 x=x12=2017



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