Library / Macedonian Mathematical Competitions /31 of 73

Macedonian Mathematical Competitions

Algebra Difficulty 4.9 AIME Prove it North Macedonia

One student was multiplying two numbers. During the multiplication he switched the last digit of the first number, which was $4$ , with $1$ . So he obtained $525$ as a result instead of $600$ . Which numbers did the student multiply?

Solution

From the condition in the problem we get that when the first number is reduced by $4-1=3$ , then their product is reduced by $600-525=75$ . So the second number is $75 \div 3 = 25$ . The first number is $600 \div 25 = 24$ . The numbers are $24$ and $25$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.