Delicious
For integers 1 ≤ m , n ≤ 1 0 0 , what is the number of ordered pairs of ( m , n ) such that 7 m + 7 n is divisible by 5?
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1 solution
I got 5000. Doesnt it should be permutation of m and n? I divided them into two sets. First set are seven with power of odd number. I take two number from here. 50 × 50 =2500
Second set are seven with power of even number I also take two number from here 50 × 50 =2500
At last i sum up all with result to 5000 can you help me checking my false?
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You can quickly test that not all pairs of odd powers will work as 7 1 + 7 1 = 1 4 = 2 ( 5 ) + 4 . Similarily, for evens 7 2 + 7 2 = 9 8 = 1 9 ( 5 ) + 3 .
Or using simple logic: ~
7
x
end with 7, 9, 3, or 1, in unit place for x=1, 2, 3, 4. Then the cycle repeats. So for 0<x<101, there will be
4
1
0
0
=
2
5
cycles. So for each of 25 values m can take, n too can take 25 values, total 25 * 25= 625. For a cycle
m
≡
1
(
m
o
d
4
)
,
p
a
i
r
s
w
i
t
h
m
≡
3
(
m
o
d
4
)
to be divisible by 10, so also with 5. So also remaining three members of one cycle of m pair with corresponding member of n. So the total choice for (m,n) is 4 * 625=~~~~
2
5
0
0
.
Note all possible values m can take and n can take has been included, and with in a cycle, there is only one value of m that that pairs with only one value of n. 7 pair with only 3, 9 pair with only 1, 3 pair with only 7, 1 pair with only 9, so as to be divisible by 5.
Brother ..7 power(4k+1)+ 7 power(4k+3) will give a number divisible by 5 and I also recognised the fact that 7 power(4k+2) + 7 power(4k) will also give the same .. Where K is a whole number
there are 25 numbers each of the form 4k+1,4k+2,4k+3 & 4k.
working out the possible pairs of odd numbered powers we get 625 (25*25) such pairs. Similarly for even number powers we get another 625. So 1250 is the answer ..
where did I go wrong? .. I am not able to figure out ..
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Don't forget the permutation.. It says the ordered pair
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Cud u pls explain my mistake ?? ... Im not able to get wat u mean
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@Ganesh Ayyappan – you count ( m , n ) = ( 4 x + 3 , 4 y + 1 ) but not ( 4 y + 1 , 4 x + 3 ) , 1 ≤ x , y ≤ 2 5 something like that...
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@Kun Pakawat – Basicly, the answer is correct, but consider that 7 1 + 7 3 and 7 3 + 7 1 are considered different since the first equation is ( m , n ) = ( 1 , 3 ) while the second equation is ( m , n ) = ( 3 , 1 )
Consider the question
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@Figel Ilham – Oh ..i thought both are the same ... So only my answer is half of original answer ... Got it!!!
I did not know about the Fermat little theorem .. By the way .. was my direction of Thinking right for this question??
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@Ganesh Ayyappan – You dont really need fermat thm here. Just consider the last digit of 7^x. It loops as 7, 9, 3, 1 then repeat.
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@Kun Pakawat – Tat was the way i thought .. 3+7 =5k
1+9=5m
M & k are positive integers
@Ganesh Ayyappan – There will be no Fermat's little theorem
By Fermat's Little Theorem , we find ϕ ( 5 ) = 4 , and we can use 7 k ≡ 2 k m o d 5 . Since 2 is a primative root mod 5, we know 2 1 ≡ 2 4 k + 1 , 2 2 ≡ 2 4 k + 2 , 2 3 ≡ 2 4 k + 3 , 2 4 ≡ 2 4 k + 4 for 0 ≤ k ≤ 2 4 , are all unique. Using this fact we find four sets ( 2 1 ) , ( 2 2 ) , ( 2 3 ) , ( 2 4 ) , each with 25 values.
Now, boiled down to 16 cases, we find only four pairs work: ( m , n ) = ( 1 , 3 ) , ( 2 , 4 ) , ( 3 , 1 ) , ( 4 , 2 ) .
Counting all possible cases we get (25)(25)(4)=2500.