1, 8, 27, 64, 124, 216, 343
Answer is 124 because in this series is cube root sequence 2 cube root is 8 ,3 cube root is 24,4 cube root is 64, 5 cube root is 125 but there is 124 which is wrong ,6 cube root is 216 ,7 cube root is 343 .
68 is the least number when divided by 8, 12, 18 and 24 leaves the remainders 4, 8, 14 and 20
385, 462, 572, 396, 427, 671, 264
427 is the answer
Because
385 means (3+5=8) or (8-3=5) ,
462 means (4+2=6) or (6-2=4) ,
572 means (5+2=7) or (7-5=2) ,
396 means (6+3=9) or (9-3=6) ,
671 means (6+1=7) or (7-6=1) ,
264 means (4+2=6) or (6-2=4) ,
but only 427 is the number in which if we do addition of any of its two digits (e.g 4+2 or 4+7) then its answer doesn’t come 3rd digit
like if we do 4+2 then answer will be 6 so the number should be 426 but it is not
or if we do 7-4 then answer will be 3 (not 2)
so the right answer is 427
4
BROTHER
Since the car has met the person 20 minutes beforehand, it has saved 10 mins of a journey
A man has started 1.30 hrs before and the car has met him 10 mins before the actual time, he takes to reach daily is 1hr and 20 mins
4/9
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
c
130 years