the digits of a certain number of 3 arithmetical digits are in arithmetical progression.Their sum is 15.And the number is 129 times the first digit.What is the number???
answer it please..
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see other post.
258 129*2= 258 2+5+8=15
To find the number, let's break down the information given step by step:
1. The number consists of 3 arithmetic digits. Let's call these digits A, B, and C in ascending order.
2. The sum of the digits is 15, so we can write the equation: A + B + C = 15.
3. The number is 129 times the first digit, so we can write the equation: ABC = 129 * A.
Now, let's solve these equations to find the values of A, B, and C.
From equation 2, we have: A + B + C = 15.
Since the digits are in an arithmetic progression, we can deduce that B = A + X and C = B + X, where X is the common difference.
Substituting the values of B and C into the equation, we have: A + (A + X) + (A + 2X) = 15.
Combining like terms, we get: 3A + 3X = 15.
Simplifying the equation, we have: A + X = 5. (Eq. 3)
Now, let's solve equation 3 for A:
A + X = 5.
A = 5 - X. (Eq. 4)
Substituting the value of A from equation 4 into equation 2, we have:
ABC = 129 * A.
(5 - X)BC = 129 * (5 - X).
Dividing both sides of the equation by (5 - X), we get:
BC = 129.
The value of BC is 129, which means B = 1 and C = 29.
Now, substitute the values of B and C back into equation 3 to find A:
A + X = 5,
A = 5 - X.
Substituting B = 1 and C = 29, we have:
A + 1 = 5,
A = 4.
So, the first digit (A) is 4, the second digit (B) is 1, and the third digit (C) is 29.
Therefore, the number is 41 and 129 times the first digit, which is 4. Multiplying, we find:
41 * 4 = 164.
Hence, the number is 164.