A natural number is said to be smart if it is divisble by the number formed by its last two digits.
For example- 3 0 2 is smart as 2 ∣ 3 0 2 .
Find the sum of the first five smart numbers having the number formed by their last two digits= 1 7 .
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Any smart number will be of form 1 0 0 x + 1 7 .
Since 1 7 ∣ ( 1 0 0 x + 1 7 ) and 1 7 ∣ 1 7 , we can say that 1 7 ∣ 1 0 0 x .
Since g cd ( 1 0 0 , 1 7 ) = 1 , we can also say that 1 7 ∣ x . Now we can solve the problem using this knowledge.
As we are looking for smallest smart numbers, value of x will be 0, 17, 34, 51 and 68. Hence the numbers are
1 0 0 ( 0 ) + 1 7 = 1 7 1 0 0 ( 1 7 ) + 1 7 = 1 7 1 7 1 0 0 ( 3 4 ) + 1 7 = 3 4 1 7 1 0 0 ( 5 1 ) + 1 7 = 5 1 1 7 1 0 0 ( 6 8 ) + 1 7 = 6 8 1 7
Sum = 17085
Note: a ∣ b means a divides b . Eg. 1 7 ∣ 3 4 and 3 ∣ 6 .
Simple standard solution.
The first number is 17 and not 117
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Thanks for pointing it out! I missed that.
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You still have a few typos. The values for x required for this problem will actually be elements of the set { 0 , 1 7 , 3 4 , 5 1 , 6 8 } . You need to fix this typo and subsequent typos because of that.
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@Prasun Biswas – Thanks for pointing it out (it was a serious typo).
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@Arulx Z – Thanks for fixing it quickly. +1 :)
P.S - There's actually a
L
A
T
E
X
command
\gcd
for the GCD notation.
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For a required smart number n we have n = 1 7 x for some natural number x .
Also n = 1 7 x ≡ 1 7 ( m o d 1 0 0 )
⇒ x ≡ 1 ( m o d 1 0 0 )
⇒ x = 1 , 1 0 1 , 2 0 1 , 3 0 1 . . . . . .
Thus the required sum is 1 7 ∗ ( 1 + 1 0 1 + 2 0 1 + 3 0 1 + 4 0 1 ) = 1 7 0 8 5 .