The gcdSumgcdSum of a positive integer is the gcdgcd of that integer with its sum of digits. Formally, gcdSum(x)=gcd(x, sum of digits of x)gcdSum(x)=gcd(x, sum of digits of x) for a positive integer xx. gcd(a,b)gcd(a,b) denotes the greatest common divisor of aa and bb — the largest integer dd such that both integers aa and bb are divisible by dd.
For example: gcdSum(762)=gcd(762,7+6+2)=gcd(762,15)=3gcdSum(762)=gcd(762,7+6+2)=gcd(762,15)=3.
Given an integer nn, find the smallest integer x≥nx≥n such that gcdSum(x)>1gcdSum(x)>1.Input
The first line of input contains one integer tt (1≤t≤104)(1≤t≤104) — the number of test cases.
Then tt lines follow, each containing a single integer nn (1≤n≤1018)(1≤n≤1018).
All test cases in one test are different.Output
Output tt lines, where the ii-th line is a single integer containing the answer to the ii-th test case.ExampleinputCopy
3 11 31 75
outputCopy
12 33 75
Note
Let us explain the three test cases in the sample.
Test case 1: n=11n=11:
gcdSum(11)=gcd(11,1+1)=gcd(11, 2)=1gcdSum(11)=gcd(11,1+1)=gcd(11, 2)=1.
gcdSum(12)=gcd(12,1+2)=gcd(12, 3)=3gcdSum(12)=gcd(12,1+2)=gcd(12, 3)=3.
So the smallest number ≥11≥11 whose gcdSumgcdSum >1>1 is 1212.
Test case 2: n=31n=31:
gcdSum(31)=gcd(31,3+1)=gcd(31, 4)=1gcdSum(31)=gcd(31,3+1)=gcd(31, 4)=1.
gcdSum(32)=gcd(32,3+2)=gcd(32, 5)=1gcdSum(32)=gcd(32,3+2)=gcd(32, 5)=1.
gcdSum(33)=gcd(33,3+3)=gcd(33, 6)=3gcdSum(33)=gcd(33,3+3)=gcd(33, 6)=3.
So the smallest number ≥31≥31 whose gcdSumgcdSum >1>1 is 3333.
Test case 3: n=75 n=75:
gcdSum(75)=gcd(75,7+5)=gcd(75, 12)=3gcdSum(75)=gcd(75,7+5)=gcd(75, 12)=3.
The gcdSumgcdSum of 7575 is already >1>1. Hence, it is the answer.
Solution: Coming Soon
Also Read
- [Solution] Christmas Game Codeforces Round #711
- [Solution] Bananas in a Microwave Codeforces Round #711
- [Solution] Two Houses Codeforces Round #711
- [Solution] Planar Reflection Codeforces Round #711
- [Solution] Box Fitting Codeforces Round #711
5 thoughts on “[Solution] GCD Sum Codeforces Round #711”