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?
let the number be
1abcdef
f=e+d
f=1+b
a=c
a=f
a+b+c+d+e+f+1=45
ok you have five equations and six unknowns. There can be no single solution.
so take the last equation, and start substituting
a + (b+1) + c+ (d+e) + f=45
f+f+f+f+f=45
f=9, a=9, c=9,b=8, or
1989de9 is the number so far...
and d+e=9
so, any combination of d and e such that the combination adds to 9 will work.
0,9
1,8
2,7,
3,4...
lets try 2,7 to see if the number satisfies the stated facts.
1989279 is between one and two million,
1989279 has one digit as the sum of tens and hundreds
1989279 has one digit as the sum of millions and ten thousands digit
1989279 has hundred thousands, thousands, and ones digit the same
1989279 has the sum of digits equal to 45
remember, any combination of d and e adding to 9 will work
To find the number that matches the given criteria, we need to break down the information provided step by step.
Let's denote the number as ABCDEF, where A represents the millions digit, B represents the hundred thousands digit, C represents the ten thousands digit, D represents the thousands digit, E represents the hundreds digit, and F represents the ones digit.
From the information given, we can deduce the following conditions:
1. The number is between 1000000 and 2000000, which implies that A = 1.
So, the number is 1BCDEF.
2. The ones digit is the sum of the tens and hundreds digits and is also the sum of the millions and ten thousands digits.
This can be expressed as F = B + E and F = A + C.
3. The hundred thousands, thousands, and ones digits are the same.
This means that B = D = F.
4. The sum of the digits is 45.
This can be expressed as A + B + C + D + E + F = 45.
Using these conditions, we can solve for the variables.
Substituting B = D = F into the equation A + B + C + D + E + F = 45, we get:
1 + B + C + B + E + B = 45
3B + C + E + 1 = 45
3B + C + E = 44 ---(Equation 1)
From the condition F = B + E, we can rewrite Equation 1 as:
3B + C + (F - B) = 44
2B + C + F = 44 ---(Equation 2)
From the condition F = A + C, we can rewrite Equation 2 as:
2B + C + (1 + C) = 44
2B + 2C + 1 = 44
2B + 2C = 43 ---(Equation 3)
Now, let's find the possible values for B and C that satisfy Equation 3.
We know that the sum of the digits is 45, so B + C + 1 + C + E + B = 45.
Reducing this equation, we get:
2B + 2C + E + 1 = 45
2B + 2C = 44
Comparing this equation with Equation 3, we can conclude that E = 0.
Now, substituting E = 0 into Equation 1, we get:
3B + C + 0 = 44
3B + C = 44 ---(Equation 4)
Next, we need to find the possible pairs of (B, C) that satisfy Equation 4.
Since the sum of the digits is 45, we can write the equation as:
B + C + F + 1 = 45
B + C + B = 45
2B + C = 45
Comparing this equation with Equation 4, we can conclude that B = 15.
Substituting B = 15 into Equation 4, we get:
3(15) + C = 44
45 + C = 44
C = -1
Since C cannot be negative, we do not have a valid solution for the given conditions.
Therefore, there is no number that satisfies all the given conditions.