i am a number less than 10.my hundreds digit is 3 more than my thousands digit,my tenths is 2 less than my hundreds and my ones is the sum of my tenths and hundreds digits.what number can i be?
n = your number
a = ones digit
b = tenths digin
c = hundreds digit
d = thousands digit
n = a + b / 10 + c / 100 + d / 1000
c = d + 3
b = c - 2 = d + 3 - 2 = d + 1
a = b + c = d + 1 + d + 3 = 2 d + 4
n = a + b / 10 + c / 100 + d / 1000
n = 2 d + 4 + ( d + 1 ) / 10 + ( d + 3 ) / 100 + d / 1000 Multiply both sides by 1000
1000 n = 1000 * ( 2 d + 4 ) + 1000 * ( d + 1 ) / 10 + 1000 * ( d + 3 ) / 100 + 1000 * d / 1000
1000 n = 1000 * ( 2 d + 4 ) + 100 * ( d + 1 ) + 10 * ( d + 3 ) + d
1000 n = 2000 d + 4000 + 100 d + 100 + 10 d + 30 + d
1000 n = 2111 d + 4130
For d = 0
1000 n = 2111 * 0 + 4130
1000 n = 0 + 4130 = 4130 Divide both sides by 1000
n = 4.130
For d = 1
1000 n = 2111 * 1 + 4130
1000 n = 2111 + 4130 = 6241 Divide both sides by 1000
n = 6.241
For d = 2
1000 n = 2111 * 2 + 4130
1000 n = 4222 + 4130 = 8352 Divide both sides by 1000
n = 8.352
For d = 3
1000 n = 2111 * 3 + 4130
1000 n = 6333 + 4130 = 10463 Divide both sides by 1000
n = 10.463
A number must be less than 10 , so that is not solution.
For d = 3 , 4 , 5 ... n > 10
The solutions are :
n = 4.130
n = 6.241
n = 8.352
To find the number that satisfies the given criteria, let's break it down step by step:
First, let's analyze the information provided. We know that:
- The number is less than 10.
- The hundreds digit is 3 more than the thousands digit.
- The tenths digit is 2 less than the hundreds digit.
- The ones digit is the sum of the tenths and hundreds digits.
Let's assign variables to the digits of the number to make it easier to work with:
- Thousands digit: Let's call it "A."
- Hundreds digit: Let's call it "B."
- Tenths digit: Let's call it "C."
- Ones digit: Let's call it "D."
Based on the given information, we can form the following equations:
1) B = A + 3
2) C = B - 2
3) D = C + B
Now, let's solve for each variable one by one:
From Equation 1, we can substitute B with (A + 3) in Equation 2:
C = (A + 3) - 2
C = A + 1
Finally, substitute C with (A + 1) in Equation 3:
D = (A + 1) + B
D = (A + 1) + (A + 3)
D = 2A + 4
Now, let's simplify the equation:
D = 2A + 4
Since the number is less than 10, we can assume that the thousands digit (A) is 0 or 1.
Case 1: A = 0
Substitute A with 0 in the equation D = 2A + 4:
D = 2(0) + 4
D = 4
Case 2: A = 1
Substitute A with 1 in the equation D = 2A + 4:
D = 2(1) + 4
D = 6
So, there are two possible numbers that satisfy the given conditions:
1) If A = 0, then the number is 401.
2) If A = 1, then the number is 611.
Therefore, the possible numbers that fit the given conditions are 401 and 611.