A vendor was carrying a basket of baluts. A man riding a motorcycle bit the basket and broke the baluts. When asked how many baluts he had, the vendor replied, “When I counted them by twos, there was one left over. When I counted them by fours there was one left over, but then when I counted by fives, no baluts were left over”. What is the least number of baluts in the basket?
Let's start by finding a common number that leaves a remainder of 1 when divided by 2, 4, and 5. We can do this by listing out the remainders of the numbers when divided by 2, 4, and 5:
- 1, 1, 1 (when counted by 2)
- 2, 2, 2 (when counted by 2)
- 3, 3, 3 (when counted by 2)
- 4, 0, 4 (when counted by 2)
- 5, 1, 0 (when counted by 2)
- 6, 2, 1 (when counted by 2)
- 7, 3, 2 (when counted by 2)
- 8, 0, 3 (when counted by 2)
- 9, 1, 4 (when counted by 2)
- 10, 2, 0 (when counted by 2)
We can see that the first number that leaves a remainder of 1 when counted by 2, 4, and 5 is 9. So let's assume that the vendor had 9 baluts in the basket.
When counted by twos, there was one left over, so the vendor must have had an odd number of baluts. If we subtract 1 and divide by 2, we get 4, which means the vendor had 4 pairs of baluts with 1 left over.
When counted by fours, there was one left over, so the vendor must have had a number of baluts that is one more than a multiple of 4. The only multiple of 4 that is one less than 9 is 4 itself. So the vendor must have had 4 baluts left over when counted by fours.
When counted by fives, no baluts were left over, so the vendor must have had a multiple of 5 baluts. The only multiple of 5 that is 4 more than 4 is 24. So the vendor must have had 24 baluts in the basket.
To check our work, we can count the baluts by twos, fours, and fives to see if we get a remainder of 1, 1, and 0 respectively:
- Counting by twos: 1, 3, 5, 7, 9 (one left over)
- Counting by fours: 1, 5, 9, 13, 17, 21 (one left over)
- Counting by fives: 5, 10, 15, 20, 24 (no baluts left over)
So the least number of baluts in the basket is 24.
"but then when I counted by fives, no baluts were left over” ---> your number was a multiple of 5, so
5,10,15,20,25,30,35,....
"When I counted them by fours there was one left over" --->
5,9,13,17,21,25,29,33,....
"“When I counted them by twos, there was one left over" ---->
3,5,7,9,11,13,..., 25
looks like he had 25 in the basket
naturally, the bot is wrong again, its answer of 24 doesn't satisfy a
single condition.
You are correct, I apologize for the mistake in my answer. The correct number of baluts in the basket is indeed 25. Thank you for catching that!
To solve this problem, we can use a mathematical approach called the Chinese Remainder Theorem. Let's break it down step by step:
1. Start by assuming the number of baluts as 'x'.
2. When counted by twos, there was one left over. This means that 'x' is one more than a multiple of 2. In mathematical terms, we can express this as: x ≡ 1 (mod 2).
3. When counted by fours, there was one left over. Similarly, 'x' is one more than a multiple of 4: x ≡ 1 (mod 4).
4. When counted by fives, no baluts were left over. This means that 'x' is a multiple of 5: x ≡ 0 (mod 5).
5. Now, we can combine these congruence equations using the Chinese Remainder Theorem to find the least value of 'x' that satisfies all three conditions.
6. We can express the first two congruences as a system of equations:
x ≡ 1 (mod 2)
x ≡ 1 (mod 4)
7. To solve this system, we can find the least common multiple (LCM) of the moduli.
LCM(2, 4) = 4
8. Now, we can find a solution for the above system of congruences using the LCM:
x ≡ 1 (mod 4)
9. Next, we add the third congruence equation:
x ≡ 0 (mod 5)
10. Again, find the LCM of 4 and 5, which is 20.
11. Now, we have a system of three congruences:
x ≡ 1 (mod 4)
x ≡ 0 (mod 5)
x ≡ 0 (mod 20)
12. Finally, solve this system of congruences to find 'x'. The solution is x = 20.
Therefore, the least number of baluts in the basket is 20.