n = a*b , numbers of the form a^b

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Source: Azerbaijan NMO 2023. Senior P1

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#1 h Aug 24, 2023, 7:43 PM

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The teacher calculates and writes on the board all the numbers $a^b$ that satisfy the condition $n = a\times b$ for the natural number $n.$ Here $a$ and $b$ are natural numbers. Is there a natural number $n$ such that each of the numbers $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ is the last digit of one of the numbers written by the teacher on the board? Justify your opinion.

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#2 h Aug 24, 2023, 9:05 PM

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If $a^b$ ends with $3$ then $2 \not |a, 2 \not |b$ so $2 \not |n=a*b$ but then $a^b$ can not ends with $2$ in such case

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#3 h Nov 23, 2024, 4:36 PM

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In order to have $7$ , our number $a^b$ should be in form of $(10k+3)^{4m+3}$ or $(10k+7)^{4m+1}$
1) $n=(10k+3)(4m+3)$ n is odd, $a^b$ cannot be even
2) $n=(10k+7)(4m+1)$ n is odd, $a^b$ cannot be even

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#4 h Apr 29, 2025, 5:31 PM

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Let us assume such a number exists. For some pair $(a, b)$ satisfying $n = ab$ , we must have $a^b \equiv 0 \pmod{10}$ .
This implies that $a$ must be even, so $n$ is even.

Also, suppose there exist $c, d$ such that $n = cd$ and $c^d \equiv 7 \pmod{10}$ .
For $c^d \equiv 7 \pmod{10}$ to hold, $c$ must be odd. Then $d$ must be even, since the product $cd = n$ is even.
Let $d = 2d_1$ . Then:
$c^d = (c^{d_1})^2 \equiv 7 \pmod{10}.$ However, for any integer $x$ , $x^2 \equiv 0, 1, 4, 5, 6,$ or $9 \pmod{10}$ . So it is impossible for $x^2 \equiv 7 \pmod{10}$ . Contradiction.

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