Maths Olympiad Prep

Track / Stage 5 / 95 of 400 #695 of 1964

Problem 695

AIME late
Number theory Difficulty 5.3 Find the answer

14. (3 points) A 1994-digit integer, where each digit is 3. When it is divided by 13, the 200th digit (counting from left to right) of the quotient is \qquad , the unit digit of the quotient is \qquad , and the remainder is \qquad .

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

【Solution】Solution: Test 3130.2307692308,33132.5384615385,3331325.61538461533333313\frac{3}{13} \approx 0.2307692308, \frac{33}{13} \approx 2.5384615385, \frac{333}{13} \approx 25.615384615 \cdots \frac{333333}{13} =25641=25641,

Therefore, the result of dividing this 1994-digit number by 13 is: 25641 in a cycle. (ignoring the decimal part), so 200÷6=332200 \div 6=33 \cdots 2,

The 200th digit (counting from left to right) of the quotient is 5;
1994÷6=3322 1994 \div 6=332 \cdots 2 \text {, }
The result of 33÷1333 \div 13 is 33÷13=2733 \div 13=2 \cdots 7,
From this, we can know that the unit digit of the quotient is 2 and the remainder is 7.
Answer: A 1994-digit number, where each digit is 3, when divided by 13, the 200th digit (counting from left to right) of the quotient is 5, the unit digit of the quotient is 2, and the remainder is 7.

Therefore, the answers are: 5,2,75, 2, 7.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.