I had a Maths doubt Let A be a six digit positive integer which is for...
To prove that the sum A + B is divisible by 91, we need to show that their sum leaves no remainder when divided by 91.
Let's start by writing A and B in terms of their digits x and y:
A = 100,000x + 10,000y + 1,000x + 100y + 10x + y
= 100,011x + 10,101y
B = 100,000y + 10,000x + 1,000y + 100x + 10y + x
= 100,001y + 10,010x + 1,100y + 100x
Now, let's calculate the sum A + B:
A + B = (100,011x + 10,101y) + (100,001y + 10,010x + 1,100y + 100x)
= (100,011x + 10,010x + 100x) + (10,101y + 100,001y + 1,100y)
= 110,121x + 111,202y
We can rewrite 110,121 and 111,202 as multiples of 91:
110,121 = 91 × 1,209
111,202 = 91 × 1,222
Now, substitute these values back into the equation:
A + B = (91 × 1,209)x + (91 × 1,222)y
= 91(1,209x + 1,222y)
Since A + B is a multiple of 91, it is divisible by 91.
Explanation:
1. Introduction:
- We need to prove that the sum A + B is divisible by 91.
2. Expressing A and B in terms of x and y:
- Write A and B in terms of their digits x and y.
3. Calculation of A + B:
- Calculate the sum A + B.
4. Expressing 110,121 and 111,202 as multiples of 91:
- Rewrite the numbers 110,121 and 111,202 as multiples of 91.
5. Substitution and simplification:
- Substitute the values back into the equation and simplify.
6. Conclusion:
- Since A + B is a multiple of 91, it is divisible by 91.