Given that:
4m+n=20; and
n≤20
How many ordered pairs (m,n) exist in which m and n both are integers?
Quant Numbers Problem
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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.
So values would be 20, 16, 12, 8, 4, 0, 4, 8, 12, 16, 20 => 11 values in total.
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
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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Solution:Neellkanth wrote: ↑Wed Apr 08, 2020 3:55 amGiven that:
4m+n=20; and
n≤20
How many ordered pairs (m,n) exist in which m and n both are integers?
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