( □ 5 ) 2 = □ 2 5
Each square above represents a positive integer. Let m and n denote the values that fill in the green and blue squares, respectively, satisfying the equation. Then what is the relationship between m and n ?
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If m is 0, n is also 0 (05^2=25) so this equation is not valid.
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It is given that each box represents a positive integer, so m = 0 is not an option. That said, if m = 0 then this yields that n = 0 ∗ ( 0 + 1 ) = 0 , and ( m , n ) = ( 0 , 0 ) does satisfy the given question as ( 0 5 ) 2 = 0 2 5 . So I believe that the equation n = m ( m + 1 ) is indeed valid for m = 0 as well, even though we weren't actually asked to consider it. :)
May I ask how you arrived at the idea to express the equation like this? I don't quite understand why/how you chose 10m and 100n. Is it because the left hand side equation has to be a two digit number and the right hand a three digit?
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This actually works for any number of digits. If for example the LHS was 2 1 5 2 then we would have m = 2 1 , for which 1 0 m + 5 = 2 1 5 as required. Similarly for the RHS, whatever positive integer n is, since it is in the hundreds position we can write n 2 5 as 1 0 0 n + 2 5 .
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To explain it more intuitively. Multiplying the number m by 10 means we are working with the set of numbers known as the "ten's". Since 5 is a unit, adding any number of tens will not change its value.
So 10 + 5 is the same as putting a 1 in front of the 5. 10m + 5, is putting ANY number in front of the 5.
m=n=0 defeats all options for m, n relations. but then yeah 0* anything = 0 solves it.
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itsnot 0x the number its 05^2 eg 5^2 = 025 i.e 25
how do U know that the number is multiplied or added inside the box....? please say me
25²=625 M=2,N=6 Use values in equation You get your solution
Yes this appears to be true for one pair of ( m , n ) . However, you need to show that it's true for all pairs of ( m , n ) .
Please review Brian's solution for a proper approach.
As the numbers represent only 1 digits, there are only 2 possibilities: 15 - 225 and 25 - 625, it is not really nice, but checking the 4 options is faster this case :)
25^(2) = 625. Hence, m=2, n=6. Only option 4 works : n = m(m+1).
Yes. This works for m = 2 , n = 6 . But is it true for all sets of ( m , n ) ? ;)
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Yup, it does. (10m+5)^2 = 100m^2+100m+25=100n+25 Hence, n=m+m^2 or n=m(m+1) Sorry for replying so late. I didn't notice the notification
How does this explain m=2 and n=4? (2 5)^2 = 4 25
I must be missing something, here. Vitor, "1"5^2=10, "2"25 = 50. So how does "1"5^2 = "2"25?
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Note that this is an arithmetic puzzle, and I explained what □ 5 means.
1
5
2
=
2
2
5
. So when
m
=
1
,
n
=
2
.
2
5
2
=
6
2
5
. So when
m
=
2
,
n
=
6
.
Ultimately, the pattern is n = m ( m + 1 ) .
Put 1 into the first one so you get 15^2 which is 225, therefore if m is 1 n is 2. Looking at the answers it has to be n=m(m+1)
That's one way to reach the answer. Can you prove that the answer is correct?
Excuse me, but if □ ∈ Z , and □ represents an order of magnitude of the decimal, positional number system, then why isn't ( 0 5 ) 2 = 0 2 5 an acceptable solution? I'm not salty, I got the right answer, but you never said that m , n = 0 .
It is asking for the general relationship, and not just a specific instance.
So, just showing that when m = 0 ⇒ n = 0 doesn't mean that all 3 of the options are correct.
In ( m 5 ) 2 = n 2 5 we have ( 1 0 m + 5 ) 2 = 1 0 0 n + 2 5 ⟹ 1 0 0 m 2 + 1 0 0 m + 2 5 = 1 0 0 n + 2 5 ⟹ 1 0 0 m 2 + 1 0 0 m = 1 0 0 n ⟹ m 2 + m = n ⟹ n = m ( m + 1 )
If n and m are both zero then 05^2 = 025 which looks right to me. That satisfies all three equations given. 0=0.
Dear David,I guess „positive integer“ does mean without 0
If I put 1 in the first square the answer will be:
"1"5² = "2"25
All the answer are: n= m ( m + x) (x could be 1, 2, 3 or other number).
So if m = 1 and n = 2
I will have:
2 = 1 ( 1 + x)
2 = 1 + x
x = 2 - 1
x = 1
So the answers will be:
n = m ( m + 1)
I used the equation 15^2 =225.
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We can write the equation as
( 1 0 m + 5 ) 2 = 1 0 0 n + 2 5 ⟹ 1 0 0 m 2 + 1 0 0 m + 2 5 = 1 0 0 n + 2 5
⟹ 1 0 0 m 2 + 1 0 0 m = 1 0 0 n ⟹ m 2 + m = n ⟹ n = m ( m + 1 ) .