In a cube all the diagonal and sides are equal, we can go diagonally.
16
324:400:576?
18×18:20×20:24×24:26×26
324:400:576:676
1972
So we consider the 2nd statement first. We can form an equation out of it.
14x-6=13y+3=9z+3
Using this, we can understand that the multiple of 14 and the multiple of 13 and 9 must have a difference of 9. The easiest way to ensure that is multiplying it by 9
14*9=126
13*9=117
If the 5th farmer gives 3 apples to the 4th farmer, they would have 123 and 120 apples respectively. However, we also know that the 2nd farmer has 117 apples (13*9=117, and this is a multiple of 9) if the 5th farmer gives 3 apples too the 2nd farmer, the 3rd, 4th and 5th farmers would have 120 apples each.
Now that we got 120, we should check if the first part of the question makes sense along with it. The equation would be
7a+1=11b-1=120
We know that 11*11=121 and 7*17=119. When we add 1 to 119 and subtract 1 from 121, we get 120 for each. In this way, all the farmers have 120 apples each.
Therefore, the 3rd farmer had a yield of 11 per tree and the 4th farmer had a yield of 9 per tree.
4 days
8400
c
To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.
The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.
There are 24 primes in S, so the product of S is:
2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97
We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.
5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.
25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.
Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros
Speed Ratio = 1:7/6 = 6:7
Time Ratio = 7:6
1 ——– 7
4 ——— ? 28 m