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Show that a perfect square cannot end with digits 2,3 or 7?

Note by Goutam Narayan 3 years, 9 months ago

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We look modulo 10.A number can give remainder 0,1,2...9 mod 10.After checking for all of these values, we see that none of them are \(\equiv\) to 2,3 or 7 (mod 10)

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can u plz show it

Well ,when a number \( n=10k+r\), then \(n^2=100k^2 + 20kr + r^2 \equiv r^2 \pmod{10}\).Now for

\(r=1, n^2\equiv 1^2 \equiv 1\pmod {10}\)

\(r=2, n^2\equiv 2^2 \equiv 4 \pmod {10}\)

\(r=3, n^2\equiv 3^2 \equiv 9 \pmod{10}\)

\(r=4, n^2\equiv 4^2 \equiv 6 \pmod{10}\)

\(r=5, n^2\equiv 5^2 \equiv 5 \pmod{10}\)

\(r=6, n^2\equiv 6^2 \equiv 6 \pmod{10}\)

\(r=7, n^2\equiv 7^2 \equiv 9 \pmod{10}\)

\(r=8, n^2\equiv 8^2 \equiv 4 \pmod{10}\)

\(r=9, n^2\equiv 9^2 \equiv 1 \pmod{10}\)

@Bogdan Simeonov – thank u very much///

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## Comments

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TopNewestWe look modulo 10.A number can give remainder 0,1,2...9 mod 10.After checking for all of these values, we see that none of them are \(\equiv\) to 2,3 or 7 (mod 10)

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can u plz show it

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Well ,when a number \( n=10k+r\), then \(n^2=100k^2 + 20kr + r^2 \equiv r^2 \pmod{10}\).Now for

\(r=1, n^2\equiv 1^2 \equiv 1\pmod {10}\)

\(r=2, n^2\equiv 2^2 \equiv 4 \pmod {10}\)

\(r=3, n^2\equiv 3^2 \equiv 9 \pmod{10}\)

\(r=4, n^2\equiv 4^2 \equiv 6 \pmod{10}\)

\(r=5, n^2\equiv 5^2 \equiv 5 \pmod{10}\)

\(r=6, n^2\equiv 6^2 \equiv 6 \pmod{10}\)

\(r=7, n^2\equiv 7^2 \equiv 9 \pmod{10}\)

\(r=8, n^2\equiv 8^2 \equiv 4 \pmod{10}\)

\(r=9, n^2\equiv 9^2 \equiv 1 \pmod{10}\)

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