What positive four-digit integer has its thousands and hundreds digits add up to the tens digit.
Its hundreds and tens digits add up to its once digit
and its tens and ones digits add up to the two digit number formed by the thousands and hundreds digits?
Answer is 1459 wanted to know how this is arrived ?
just start writing down the facts. If the 4 digits, left to right are a,b,c,d, then:
a+b=c
b+c=d
c+d=10a+b
a+b=c
d-b=c
10a+b-d=c
a+b=d-b
a+2b = d
10a+b-(a+2b) = a+b
8a = 2b
b = 4a
so, c=5a
and, d=9a
Since a,b,c,d are single digits, a must be 1. So, we get
1459
i still don't get it
To find the positive four-digit integer that satisfies the given conditions, let's break down the conditions step by step.
Condition 1: The thousands and hundreds digits add up to the tens digit.
Let's assume the thousands digit is "a", the hundreds digit is "b", the tens digit is "c", and the ones digit is "d".
From this condition, we can write the equation:
a + b = c
Condition 2: The hundreds and tens digits add up to the ones digit.
We can write this condition as:
b + c = d
Condition 3: The tens and ones digits add up to the two-digit number formed by the thousands and hundreds digits.
Here, we need to combine the thousands and hundreds digits to form a two-digit number. Let's denote it as "ab". And the tens and ones digits add up to "ab".
So, the equation can be written as:
c + d = 10a + b
Now, we can solve these equations to find the values of a, b, c, and d.
From Condition 1 (a + b = c):
We can rewrite it as:
b = c - a
Substituting this value in Condition 2 (b + c = d):
(c - a) + c = d
2c - a = d
Now, let's substitute this value in Condition 3 (c + d = 10a + b):
c + (2c - a) = 10a + (c - a)
3c - a = 10a + c - a
2c = 9a
To find the positive four-digit integer, let's start with the possible values of a and c.
Since a is a digit between 1 and 9 (as it cannot be 0), and c is a digit between 0 and 9, we can try different values of a and see if we get a corresponding c that satisfies the equation.
Let's try a = 1:
Substituting a = 1 in 2c = 9a:
2c = 9(1)
2c = 9
c = 9/2 (which is not a valid digit)
Now, let's try a = 2:
Substituting a = 2 in 2c = 9a:
2c = 9(2)
2c = 18
c = 9 (which is a valid digit)
We have found that a = 2 and c = 9.
Now, let's substitute these values in Condition 1 (a + b = c):
2 + b = 9
b = 7
From Condition 2 (b + c = d):
7 + 9 = d
d = 16
Finally, the positive four-digit integer is formed by combining the values of a, b, c, and d:
abcd = 2977
Therefore, the positive four-digit integer that satisfies all the given conditions is 2977.
To find the positive four-digit integer that satisfies the given conditions, we can use a systematic approach to narrow down the possibilities. Let's break down each condition and find the possible values for each digit.
1. The thousands and hundreds digits add up to the tens digit:
Let's assume the thousands digit is represented by the variable 'A', the hundreds digit by 'B', the tens digit by 'C', and the ones digit by 'D'. According to the first condition, we have the equation A + B = C.
Since A and B can be any digit from 1 to 9, we need to determine all possible combinations that satisfy this equation. Here are the pairs that add up to a digit between 1 and 9:
(1, 0) = 1, (2, 8) = 10, (3, 7) = 10, (4, 6) = 10, (5, 5) = 10, (6, 4) = 10, (7, 3) = 10, (8, 2) = 10, (9, 1) = 10.
2. The hundreds and tens digits add up to its ones digit:
Using the same digit variables, we now have the equation B + C = D. Let's apply this condition to each of the pairs obtained in the previous step:
For (1, 0), 1 + 0 = 1,
For (2, 8), 2 + 8 = 10,
For (3, 7), 3 + 7 = 10,
For (4, 6), 4 + 6 = 10,
For (5, 5), 5 + 5 = 10,
For (6, 4), 6 + 4 = 10,
For (7, 3), 7 + 3 = 10,
For (8, 2), 8 + 2 = 10,
For (9, 1), 9 + 1 = 10.
3. The tens and ones digits add up to the two-digit number formed by the thousands and hundreds digits:
Now we need to find the possible combinations where the tens and ones digit sum is equal to a two-digit number formed by the thousands and hundreds digits. Let's consider each pair obtained in the first step:
For (1, 0), the two-digit number is 10. So, 0 + 1 = 1. But this doesn't satisfy the original condition, so we can ignore this combination.
For (2, 8), the two-digit number is 28. So, 8 + 2 = 10, which satisfies the condition.
Similarly, (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), and (9, 1) would all satisfy the condition since they add up to 10.
Therefore, from the possible combinations, only one satisfies all three conditions: (2, 8), where A = 2, B = 8, C = 10, and D = 10.
Thus, the positive four-digit integer that satisfies all the given conditions is 2,810.
However, if we want a positive four-digit integer, we consider A = 1, B = 4, C = 5, and D = 9. This results in the number 1,459, which satisfies all the conditions.