Question

if the digits of integer x are reversed and the resulting number is added to the original x the sum is 7777 what is the smallest possible value of x?

Answer 1

Let the integer x be:

ijkl

When we reverse it we have:

lkji

The sum of the two would give:

l + i k + j j + k i + l

The possible sums of 7 are :

4 + 3

2 + 5

1 + 6

0 + 7

the first digit will not be zero.

We could have the smallest x having the smallest digits for the sum of 7.

The smallest first digit we could have is 1 and thus the last digit will be 6.

The smallest second number we could have is 0 and thus the third number could be 7

The smallest x is as follows :

i = 1

j = 0

k = 7

l = 6

The number is thus :

= 1076

Answer 2

$\textbf {Hey there, here is the solution.}$

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Define the number:

Let the number be abcd.

The reverse of the number will be dcba.

.

The sum of the number and its reverse is 7777 (Given)

⇒ abcd + dcba = 7777

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STEP 1 : Form equation:

a + d = 7

b + c = 7

c + b = 7
(duplicate)

d + a = 7
(duplicate)

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STEP 2: Find the value of a:

Since "a" has the biggest place value, "a" should be smallest possible,

⇒ a = 1 (since "a" is the first digit , it cannot be zero)

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STEP 3: Find the value of d:

a + d = 7

d = 7 - 1

d = 6

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STEP 4: Find the value of b:

zero is the smallest possible digit.

Smallest possible value of b = 0

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STEP 5: Find the value of c:

b + c = 7

c = 7 - 0

c = 7

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STEP 6: Find the number:

Put all the digit together.

abcd = 1076

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Answer: The number is 1076

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⭐$\textbf {Cheers}$⭐