Question

The Sentosa High School's telephone number is an eight digit number. The sum of the two numbers formed from the first three digits and the last five digits respectively 66558. The sum of two numbers formed from the first five digits and the last three digits is 65577. Find the telephone number of the Senthosa High School.

Answer 1

Let's the phone number is xabycdef. Because we have to sum two pairs of 5-digits and 3-digits number. So it can conducted that x=6 and y=6. We have:

6cdef 6ab6c
6ab def
------- --------
66558 65577
From the first sum, d + 6 = ..5 => c + 1 = 6 => c=5
From the last digits in second sum: c + f = 5 + f = 7 => f = 2
Similarly, f + b = 8 => b=6
From the second sum: b + d = ..5 => a + 1 = 5 => a = 4
Go further, you will find the phone number is 64665912

Answer 2

64665912

Answer 3

How did you get the answer?

Answer 4

hey, d and e = ?

Answer 5

[AI-generated answer — Explain Bot]

To find the telephone number of Sentosa High School, we need to break down the given information and solve for each digit step by step.

Let's say the telephone number of Sentosa High School has the following format: ABCDEFGH, where each letter represents a digit.

According to the problem, the sum of the two numbers formed from the first three digits (ABC) and the last five digits (DEFGH) is 66558. This can be written as:

(100 * A + 10 * B + C) + (10000 * D + 1000 * E + 100 * F + 10 * G + H) = 66558

Simplifying the equation gives us:

100A + 10B + C + 10000D + 1000E + 100F + 10G + H = 66558

Similarly, the sum of the two numbers formed from the first five digits (ABCDE) and the last three digits (FGH) is 65577:

(10000 * A + 1000 * B + 100 * C + 10 * D + E) + (100 * F + 10 * G + H) = 65577

Simplifying the equation gives us:

10000A + 1000B + 100C + 10D + E + 100F + 10G + H = 65577

Now we have a system of two equations with eight variables. To solve this system and find the telephone number, we can use the method of substitution or elimination.

Let's solve it using the substitution method:

From the first equation, we can express C in terms of the other variables:

C = 66558 - 100A - 10B - 10000D - 1000E - 100F - 10G - H

Now we substitute this expression for C in the second equation:

10000A + 1000B + 100(66558 - 100A - 10B - 10000D - 1000E - 100F - 10G - H) + 10D + E + 100F + 10G + H = 65577

Now simplify this equation:

10000A + 1000B + 6655800 - 10000A - 1000B - 1000000D - 100000E - 10000F - 1000G - 100H + 10D + E + 100F + 10G + H = 65577

Combine like terms:

-9999000D - 999000E - 99900F - 9990G - 999H + 10D + E + 6555800 = 65577

Combine constants:

-9999000D - 999000E - 99900F - 9990G - 999H + 10D + E = 65577 - 6555800

Simplify:

-9998980D - 998999E - 99888F - 9989G - 998H = -6480223

Now we have eliminated C and H from the equation. We can repeat this process of elimination for the remaining variables and solve for A, B, D, E, F, and G.