The value of a 2 digit number is 4 less than 6 times the sum of its digits.?

Question

If the digits of the number are reversed, the resulting number is 9 less than the original number. Find the number.

Answer 1

2 digit # is 10x+y
10x+y=(x+y)*6-4
10x+y=6x+6y-4
4x-5y=-4

10x+y-(10y+x)=9
9x-9y=9
x-y=1

4x-5y=-4
x-y=1 multiply both sides by -4
-4x+4y=-4
4x-5y=-4 add
-y=-8
y=8
x-y=1
x-8=1
x=9
the number is 98
check
(9+8)*6-4=17*6-4=102-4=98

Answer 2

let assujme numbet is in the form of 10x+y

First condition - If the digits of the number are reversed, the resulting number is 9 less than the original number.

10x + y - (10y + x)=9
9x - 9y = 9
x - y = 1
x = y + 1

Possible combination of numbers are
xy
10
21
32
43
54
65
76
87
98

Second condition is =The value of a 2 digit number is 4 less than 6 times the sum of its digits

10x + y = 6 (x + y) -4
10x + y = 6x + 6y -4
4x + 4 = 5y
y=4(x + 1)/5

Since Y is interger, then (x+1)/5 should be 1, 2, 3
Since Y should be less than <=9 (for two digit numbers), only possible combination are (x+1)/5= 1 or 2
For (x+1)/5=1 means x is 4 and y is 4, but it does not satisfy condition 1.
For (x+1)/5=2 means x is 9 and y is 8, which does satisfy condition 1.

So answer is 98.
Bye

Answer 3

So, let's call your original two-digit number (in the form AB) X. X = (A * 10) + B, because you have A in the tens place and B in the ones place.
You're told X = (A*10) + B = (6 * A+B) - 4
And (B*10) + A = X - 9 = (A*10) + B - 9

So, equation 1, 10A + B = 6A + 6B - 4
And equation 2, 10B + A = 10A + B - 9

From now on, it's just solving a system of two linear equations to find the two digits of the original number.

Isolate A in equation 2: 10A - A = 10B - B + 9
A = (9B + 9) / 9
A = B+1
And substitute into 1: 10(B+1) + B = 6(B+1) + 6B -4
10B + 10 + B = 6B + 6 + 6B - 4
11B + 10 = 12B + 2
8 = B
And so, A = B+1 = 9
So, our original two-digit number is 98.

To check:
98 = 6(9+8) - 4
= 6(17) - 4
= 102 - 4
= 98
and
89 = 98 - 9

Answer 4

Let the first digit be "m" and the second "n". The number is then 10m+n. The sum of the digits is m+n. Therefore,

10m+n=6(m+n)-4.

Also, the number resulting from reversing the digits is 10n+m, so

10n + m = 10m + n - 9.

if we solve each of these equations for n, we get

n = (4m + 4)/5

and

n = (9m - 9)/9 = m-1

so now we have two expressions for n, and they must be equal:

m-1 = (4m + 4)/5.

solving for m, we get:

m = 9.

Since

n = m - 1

We also have

n = 9 - 1 = 8.

Therefore the first digit is 9 and the second 8, so the number is 98.

Answer 5

98 is the answer
6(x+y)-4=10x+y
6x+6y-10x-y=4
5y-4x=4
y=(4+4x)/5
when x=9, y=8
and when x=4, y=4
both 98 and 44 are the two digit numbers which is 4 less than 6 times the sum of its digits.
but, since the numbers are reversed, resulting number is 9 less than the original number, the answer is 98.

Answer 6

Let the digits be a and b therefore number is 10a +b
sum of digits = a + b, 6 times of this = 6(a+b)
Therefore from 1st statement 10a + b = 6(a+b) – 4
solving this 10 a + b = 6a + 6b – 4
4a– 5b = – 4 --------------eq. 1
The digits in the reversed number will be b and a therefore number will be 10b +a
Therefore from 2nd statement 10b + a = 10a+b – 9
solving this 9 b – 9a = – 9 divide this eq. by 9
Therefore b – a = – 1
or a – b = 1 ---------------eq. 2 multiply this eq. by 4
4a – 4b = 4 ----------eq. 3
4a– 5b = – 4 --------------eq. 1
subtract eq. 1 from 3
b = 8
putting the value of b = 9 in eq. 1
a = 9
Hence the number is 98.

Answer 7

Let a be the "tens digit" and b the "units" digit.
Its value is 10a+b
Then 6 (a+b) - 4 = 10a+b
or 4a-5b=-4
Reversing digits 10b+a+9=10a+b
or 9a-9b = 9
Thus, the number is 98

Answer 8

98

Answer 9

98

Answer 10

98

Answer 11

10x + y = 6(x + y) - 4
10x + y = 6x + 6y - 4
4x - 5y = -4

10y + x = 10x + y - 9
9y - 9x = -9
y - x = -1
y = x - 1

4x - 5(x - 1) = -4
4x - 5x + 5 = -4
-x = -9
x = 9

y = 9 - 1
y = 8

If i have this correct, the answer is 98.