Method
{"text":"bold","word":"The Approach 1"}
Part (b): Finding the Four-Digit Number
ABCD
We are tasked to find a four-digit number
ABCD, such that when multiplied by 9, the result is the reverse of the original number,
DCBA. In other words:
ABCD×9=DCBA
Steps to Solve:
Express the problem mathematically:
Let
ABCD=1000A+100B+10C+D.
Let
DCBA=1000D+100C+10B+A.
The equation becomes:
9×(1000A+100B+10C+D)=1000D+100C+10B+A
Simplify the equation:
Expand both sides:
9000A+900B+90C+9D=1000D+100C+10B+A
Rearrange terms:
8999A+890B?10C?991D=0
Trial and Error with Logical Constraints:
D are digits between 1 and 9 (since
ABCD is a four-digit number).
ABCD×9=DCBA, so the result must still be a four-digit number, implying
ABCD?1000 and
ABCD<1112 (since
1112×9=10008, which is no longer four digits).
Testing Values:
Try
ABCD=1089:
1089×9=9801
The result, 9801, is the reverse of 1089.
Final Answer for Part (b):
The four-digit number is:
1089?
{"text":"bold","word":"The Approach 2"}
Work with the digits ABCD x 9 = DCBA. Replace the number 9 with
(10 - 1) and work your way from there, subtracting digits in the ones column, tens column and hundreds column and then solving by substituting.
{"text":"bold","word":"The Solution"}
{"text":"bold","word":"Step 1"}
ABCD x 9 = DCBA
ABCD (10 - 1) = DCBA
ABCD0 - ABCD = DCBA
(Because for example, 5 x 10 = 50. So the 0 comes in the one's place of the result.
In other words, we can write the equation as
ABCD0
-ABCD
---------
DCBA
---------
{"text":"bold","word":"Step 2"}
The last digit 0 is the smallest whole number and we cannot subtract D(the D in ABCD - the bottom number) from 0 without borrowing from the digit D in the tens place of ABCD0. The digit 0 borrows 1 ten from D (in ABCD0) then 0 becomes 10 and the D in ABCD0 becomes (D-1).
In the ones column of digits, 10 - D = A
Adding D on both sides,
10 - D + D = A + D
10 = A + D...............(1)
{"text":"bold","word":"Step 3"}
In the tens column of digit, (D - 1) - C = B
Adding 1 on both sides of the equation in order to move the 1 to the right-hand side of the equation.
D - 1 + 1 - C = B + 1
Subtract B on both sides to bring all the letters together and keep the 1 on the right-hand side.
D - C - B = B - B + 1
D - C - B = 1...........(2)
{"text":"bold","word":"Step 4"}
In the hundreds column of digits, C - B = C
The two Cs cancel out and we get
- B = 0
Now add B on both sides in order to make - B, positive.
- B + B= 0 + B
0 = B
or B = 0....................(3)
{"text":"bold","word":"Step 5"}
In the thousands column of digits, AB - A = D
Substitute (3) in this equation.
A0 - A = D
A0 = A x 10
So 10A - A = D
A (10 - 1) = D
9A = D.....................(4)
{"text":"bold","word":"Step 6"}
From (1) and (4), we get,
A + 9A = 10
10A = 10
Divide both sides by 10 to isolate A.
10A/10 = 10/10
A = 1......................(5)
{"text":"bold","word":"Step 7"}
From (4) and (5) we have,
9A = D
9 x 1 = D
D = 9........................(6)
{"text":"bold","word":"Step 8"}
Substitute (3) and (6) in (2) to get C,
D - C - B = 1
D - C - 0 = 1
D - C = 1
9 - C = 1
Add C on both sides to move -C to the right-hand side and make it positive.
9 - C + C = 1 + C
9 = 1 + C
Subtract 1 on both sides.
9 - 1 = 1 - 1 + C
8 = C.....................(7)
We now have all the digits,
ABCD = 1089
Let's verify it.
1089 x 9 = 9801
