Maths Olympiad Prep

Track / Stage 5 / 138 of 400 #738 of 1964

Problem 738

AIME late
Number theory Difficulty 5.3 Find the answer

(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. Fractions can be typed as 3/2, and spacing doesn't matter.

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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 reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.