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Can someone solve this?
A and B have some chocolates with them. A told B, " If you give me ’x’ chocolates both of us will have equal number of chocolates. B then told A if you give me thrice as many chocolates I would have 80 more chocolates than you, find ’x’

The equations look something like (A+x) =(B-x) and (B+3x)-(A-3x)=80

So x=10?

the-median-of-a-set-of-5-different-numbers-is-x-263613.html

please explain this question in a steps, if someone can

The exponent of a^x is a multiple of 20, what is the minimum possible value of (a + x), if a is a factor of x? (a & x are positive integers)

A. 10

B. 15

C. 20

D. 40

E. 50

it’s 800 level ques..

To solve the problem, we need to find positive integers a and x such that:

1. a^x has an exponent that is a multiple of 20.
2. a is a factor of x.
3. We want to minimize a + x.

Since the exponent of a^x must be a multiple of 20, x itself must be a multiple of 20. Let’s assume x = 20k for some integer k, and a must be a factor of x.

Let’s check small values for k to find the minimum a + x:

Case 1: k = 1
- x = 20
- a must be a factor of 20.
The factors of 20 are 1, 2, 4, 5, 10, and 20.

Calculate a + x for each factor:
1. a = 1: a + x = 1 + 20 = 21
2. a = 2: a + x = 2 + 20 = 22
3. a = 4: a + x = 4 + 20 = 24
4. a = 5: a + x = 5 + 20 = 25
5. a = 10: a + x = 10 + 20 = 30
6. a = 20: a + x = 20 + 20 = 40

The smallest value of a + x in this case is 21.

Case 2: k = 2
- x = 40
- a must be a factor of 40.
The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40.

Calculate a + x for each factor:
1. a = 1: a + x = 1 + 40 = 41
2. a = 2: a + x = 2 + 40 = 42
3. a = 4: a + x = 4 + 40 = 44
4. a = 5: a + x = 5 + 40 = 45
5. a = 8: a + x = 8 + 40 = 48
6. a = 10: a + x = 10 + 40 = 50
7. a = 20: a + x = 20 + 40 = 60
8. a = 40: a + x = 40 + 40 = 80

The smallest value of a + x in this case is 41.

Checking k > 2 will increase x and thus increase a + x. Therefore, the minimum value of a + x is obtained for k = 1 and a = 20, which gives:

a + x = 20 + 20 = 40

Thus, the minimum possible value of a + x is:
40
IMO

­This question is discussed here:

https://gmatclub.com/forum/the-exponent ... 67355.html

Hope it helps.

Where did u get the question

Bunuel, i don’t understand the explanation there all seems similar , if possible could you explain in a simple way in steps

nitish2610, how you got

nitish2610, thanks waiting....

Since the factors of 20 = 2^2 x 5 therefore for a^x to be a multiple of 20, following cases for ‘a’ can be made:
1. a = 20 with x being at least 1
2. a = 10 with x being at least 2
We also know that ‘a’ is a factor of ‘x’ (OR ‘x’ is a multiple of ‘a’) and (a+x) is to be minimized.
Since we have already considered the smallest value of ‘a’ possible, only way to pick smallest value of ‘x’ is to pick the smallest multiple of ‘a’ which will be ‘a’ itself (smallest multiple of any number will be the number itself). Therefore:
1. If a=20 then smallest value of ‘x’ will be 20 which gives (a+x) = 40
2. a=10 then the smallest value of ‘x’ will be 10 which gives (a+x) = 20

If one missed the math, some logic and options can come to rescue!

20 has factors : 1,2,4,5,10,20

20 multiplied something will always end with a zero.

2,4,5 to the power any integer is never going to end on a zero.

So, “a” can be 1, 10 or 20.

“a” is a factor of “x”, so “a” is lower or at max equal to “x” (‘cuz that’s what a factor is )

Now look at the options,
if “a” is 1, 1 raised to anything will not be a “multiple of 20”

if “a” is 10,
x = 0 for a + x = 10 (Option A) makes no sense,
x = 5 for a + x = 15 also makes no sense as “a” must be lower or at max equal to “x”.

x = 10 for a + x = 20 fits both the conditions (10 is a multiple of 10 AND 10^10 will be a multiple of 20). So, as of now, a + x minimum can be 20.

All other options for a + x are greater than 20. So we don’t need to calculate them, thus making our Answer (C) 20.

Could have eliminated 1 as well, alongside 2,4, and 5.

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