A number has three digits and is equal to 12 times the sum of its digits what the number is?
Assume three digit number = xyz, then
100x + 10y + z = 12(x + y + Z)
100x + 10y + z = 12x + 12y + 12Z
Solve to get
88x = 2y + 11z
Take x=1, then y=0 and z= 8 in order to satisfy the equation.
Three digit number is 108, hence 108 = 12(1+0+8).
Is there any other way to solve this problem?
The sum of first N integers is a three digit number with all of its digits equal, what is the value of N?
1,2,3,4,5,6,7……………. Is an arithmetic series
S= n/2[2a + (n-1) d] = n (n +1) / 2
n (n +1) / 2 = 111x
Since sum of first integers is a three digit number with all digits equal
Hence it xxx = 100x +10x +x = 111x.
By trial and taking x=6, and solving for n, n2 +n – 1332 = 0, I will get n=36, is there any other way of solving;
n (n +1) / 2 = mod 111, and getting value of n=36 from this equation?