I am a number between 1000000 and 2000000. My ones digit is the sum of my tens and hundreds digits and is also the sum of my millions and ten thousands digits.my hundred thousands, thousands, and ones digits are the same.the sum of my digits is 45.what number am i?
I am a number between 1000000 and 2000000
---> 1 ??? ???
my hundred thousands, thousands, and ones digits are the same
---> 1 x?x ??x
---> 1 xax bcx
My ones digit is the sum of my tens and hundreds digits
---> x = b+c
also the sum of my millions and ten thousands digits
---> x = 1 + a
so we know
a+1= b+c
a = b+c - 1
the sum of my digits is 45
1 + 3x + a + b + c = 45
1 + 3(1+a) + a + b + c = 45
1+3+3a+a+b+c = 45
4a + b + c = 41
4(b+c-1) + b + c = 41
4b + 4c - 4 + b + c = 41
5b + 5c = 45
b+c = 9
c = 9-b
so we have :
c = 9-b
a = b+c-1
x= a+1
let b = 0, c = 9 ,a = 8 , x = 9 --> 1,989,099
let b = 1, c = 8, a = 8, x = 9 ---> 1,989,189
let b = 2, c = 7, a = 8, x = 9 ---> 1,989,279
...
let b = 8, c = 1, a = 8 , x = 9 ---> ..........
let b = 0 , c = 9, a = 10 not possible
To find the number that satisfies all the given conditions, let's break down the problem step by step:
1. Start with the fact that the hundreds, tens, and ones digits are the same. We can represent this as "AAA000A."
2. We know that the sum of the digits is 45. Since all the hundreds, tens, and ones digits are the same, their sum will be 3 times that digit. Therefore, "3A0+3A = 45."
3. Simplify the equation: "6A = 45." Solve for A by dividing both sides of the equation by 6: "A = 45/6 = 7.5." However, since all the digits should be whole numbers, we can conclude that A must be 8.
4. Now that we have the value of A, we can substitute it back into our initial representation: "8 8 8 0008."
5. According to the problem statement, the ones digit is the sum of the tens and hundreds digits and also the sum of the millions and ten-thousands digits. So, 8 = 8 + 8 and 8 = 0 + 0. These conditions hold, confirming that our number is correct.
Therefore, the number you are is 888,008.