Number Theory (SAT)

Since out of 30 students 36 play games; the least number of students who play both games is 6. So, it is (4) that is not possible. (Since it is not given that every student plays at least one game, there could be instances like 16 play both games, and 4 play only football; Similarly other choices can be verified.)

Since 10² = 100, 14² = 196, there are 5 integers (10 to 14) whose square lies between 100 and 200.

Assume X as 3. Then (1) = 8, (2) = 21, (3) = 33, (4) = 29 and (5) = 14. So (1) and (5) is not the answer. Let take X = 2 then (2) = 11, (3) = 18, and (4) = 14. Only choice (2) is odd in both cases.

Choice (3) i.e 30 is divisible by both 15 and 10 and is least positive integer. 15 is not divisible by 10 and 20 is not divisible by 15. Other choices are more than 30 hence can be eliminated

If the division of 96 by m leaves the remainder 5, then 91 must be divisible by m without remainder.  Similarly, if the division of 120 by m leaves the remainder 3, then 117 must be divisible by m without  remainder. So, m must be a divisor of both 91 and 117. Among the choices, it is only 13 which satisfies this  condition.

Let three consecutive positive even integers are (a – 2), a, (a + 2), their sum is 3a. As a is the middle number and is an even integer. So the sum must be divisible by 6. Out of given choices, choice (3) is not a multiple of 6.

[- 2.6] - [2.6] = - 3 - 2 = - 5

Numbers divisible by both 6 and 10 are 30, 60,90,120, .... All these are not divisible by 12.

We should try one odd value (say, 1) and one even value (say 2) for n for each of the choices. Let us start with choice (5). lf n = 1, (5) = 1+ 1 + 1 = 3. If n = 2, (5) = 4 + 2 + 1 = 7. Both are odd numbers. So, we can stop here and choose (5) itself as the answer. (You can independently verify why  the other choices are wrong.).

LCM of 3, 4 ,5 and 6, which is equal to 60. (You can also argue that 30 is not divisible by 4, 40 is not divisible by 3, and 60 is divisible by all the given numbers and must be the answer.)

The possible values for (m, n) consistent with the given conditions are (4, 3 ). Then m – n = 1 or 7. So, the maximum value is 7.

Any number that is a multiple of both 3 and 5 must be a multiple of 15. The number of integers from 1 and 900 which are multiples of 15 are 15 × 1, 15 × 2, 15 × 3, ..., 15 × 60. These are 60 in number. So, the probability that one among the first 900 numbers chosen at random being a multiple of 15 is 60/900 = 6/90 = 1/15 = 3/45

Let three consecutive positive integers are (a – 1), a, (a + 1), their sum is 3a. As a is the middle number and is an integer. So the sum must be divisible by 3. Out of given choices only choice (1) is a multiple of 3.

Let (a, b) be (-1, -2). Then (1) = 2, (2) = - 2, (3) = - 3, (4) = 5 and (5) = - 7. So (4) is greatest among all the choices, is the answer.

Let a = 3 and b = 2. Then, starting from (5), the choices are (4 – 3 = 1), (9 – 2 = 7), (6 – 1 = 5), (3 x 2² = 12) and {3(3 + 2) = 15}. Of these, only the second choice  is an even integer.

Correct Answer: 0

Consistent with the given condition (a, b, c) can be 1/2, 1/3 and 6. So, I need not be true. Similarly, (a, b, c) can be (1/2, 1/2, 2). So, II need not be true. If ac and bc are both integers, then (ac)² + (bc)² must also be an integer. So, III must always be true. So, (3) is the answer.

Since we have to test all the choices, Let us consider one odd value (1) for k. Then (1) = 1, (2) = 0 , (3) = 0, (4) = 4 and (5) = 7. Now let us consider one even value (2) for k. Then (1) = 4, (2) = 2, (3) = 1, (4) = 7 and (5) = 11.In both cases Choice (2) is even, is the answer.

Obviously 98,665 is 100 less than 98, 765.

Let 5 consecutive integers are X - 2, X - 1, X , X + 1 and X + 2. Their sum is 5X. So 5X= 350 which means X is 70 which is the middle number.

The expression is of the form p²X. So, whenever p is even, the product will be even, irrespective of whether the second number X is odd or even. So, (1) is the answer.

Let us try two pairs of consecutive positive integers starting with an odd number and an even number, say (1, 2) and (2, 3). The differences between the cubes of the two numbers are (8 - 1) = 7 in  the first case, and (27 - 8) = 15 in the second case. Among the choices, it is only (2) which is true in both these cases.

Choices 1, 2, 4 and 5 can be even or odd depending on the value chosen for n.
Now, n³ - n = n(n² -1) = n(n +1)(n - 1) = (n - 1)(n)(n + 1).
This is the product of 3 consecutive integers at least one of which will be even.
Hence, the product will always be even.
Hence, choice (3) wil always be even.

p; that can be written using these three digits is 789. Their difference is 987 - 789 = 198.