Câu hỏi của Nguyễn Đức Thiện

Question

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Em xin chân thành cảm ơn.

given the sequence : 2,7,12,17,22,27,...,302,307.Find the unit digit of the product of all terms in the sequence above.

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Lê Song Phương · Accepted Answer

Dịch đề: Cho dãy số 2, 7, 12, 17, 22, 27,..., 302, 307. Tìm chữ sồ hàng đơn vị của tích tất cả các số trong dãy.

Solution (in English):

In the sequence, there are $\left(302-2\right):10+1=31$ numbers that end with 2 and $\left(307-7\right):10+1=31$ numbers that end with 7.

Therefore, the unit digit of the product is also the unit digit of:

$A=2\times2\times2\times...\times2\times7\times7\times7\times...\times7$

(31 numbers 2) (31 numbers 7)

We have to find out what is the final digit of $2\times2\times...\times2$ (31 times) and $7\times7\times...\times7$ (31 times)

Consider the number $N=2\times2\times...\times2$ ($n$ times)

If $n=1$, then $N=2$

If $n=2$, then $N=4$

If $n=3$, then $N=8$

If $n=4$, then $N=16\rightarrow$ the last digit is 6

If $n=5$, then $N=32\rightarrow$ the last digit is 2

If $n=6$, then $N=64\rightarrow$ the last digit is 4

If $n=7$, then $N=128\rightarrow$ the last digit is 8.

If $n=8$, then $N=256\rightarrow$ the last digit is 6.

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From here we can see a pattern: If $n$ has a remainder of 3 when being divided by 4, the last digit of N is 8. Thus, the final digit of $2\times2\times...\times2$ (31 times) is 8 since 31 has a remainder of 3 when being divided by 4.

Using the same method of reasoning, we can deduce that $7\times7\times...\times7$ (31 times) ends with 3.

Therefore, the final digit of A must be the final digit of $8\times3=24$, which is 4.

So the answer is 4.

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