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# Number theory

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

Note by Goutam Narayan
3 years, 6 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) · 3 years, 6 months ago

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can u plz show it · 3 years, 6 months ago

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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

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

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

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

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

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

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

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

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

$$r=9, n^2\equiv 9^2 \equiv 1 \pmod{10}$$ · 3 years, 6 months ago

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thank u very much/// · 3 years, 6 months ago

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