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Quant- Numbers Problem

Problem Solving — algebra and arithmetic
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Source: — Quantitative Reasoning |
Junior | Next Rank: 30 Posts
Posts: 15
Joined: Sat May 30, 2020 10:48 pm

Re: Quant- Numbers Problem

by orthodoxparadox » Sat May 30, 2020 10:57 pm
Let us try it for all n from -20 to 20. We can only choose such n that 20 - n is divisible by 4, since m is also an integer. Since 20 is divisible by 4, n must also be divisible by 4.

So values would be -20, -16, -12, -8, -4, 0, 4, 8, 12, 16, 20 => 11 values in total.
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Junior | Next Rank: 30 Posts
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Re: Quant- Numbers Problem

by Ignite » Sat Aug 01, 2020 10:19 pm
4m+n=20; and |n|≤20

When,
m = 0, n = 20
m = 1, n = 16
m = 2, n = 12
m = 3, n= 8
m= 4, n =4
m=5, n = 0

But, n can also take negative values upto -20
So there will be
m= 6, n = -4
m=7, n = -8
m= 8, n=-12
m=9, n =-16
m=10, n =-20

Thus, there are 11 ordered pairs.

Thanks,
Ignite
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Re: Quant- Numbers Problem

by Scott@TargetTestPrep » Tue Jul 12, 2022 1:23 am
Neellkanth wrote: ↑
Wed Apr 08, 2020 3:55 am
 Given that:

4m+n=20; and
|n|≤20

How many ordered pairs (m,n) exist in which m and n both are integers?
Solution:

We see that:

m = (20 - n)/4 = 5 - n/4 = integer

Since n is also an integer, n must be a multiple of 4. Since |n| ≤ 20, n can be 20, 16, 12, 8, 4, 0, -4, -8, -12, -16, or -20. Since there are 11 values of n, there are 11 ordered pairs (m, n).

Answer: 11
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