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प्रश्न
If from a two-digit number, we subtract the number formed by reversing its digits then the result so obtained is a perfect cube. How many such numbers are possible? Write all of them.
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उत्तर
Let ab be any two-digit number.
Then, the digit formed by reversing it digits is ba.
Now, ab – ba = (10a + b) – (10b + a)
= (10a – a) + (b – 10b)
= 9a – 9b
= 9(a – b)
Further, since ab – ba is a perfect cube and is a multiple of 9.
∴ The possible value of a – b is 3.
i.e. a = b + 3
Here, b can take value from 0 to 6.
Hence, possible numbers are as follow.
For b = 0, a = 3, i.e. 30
For b = 1, a = 4, i.e. 41
For b = 2, a = 5, i.e. 52
For b = 3, a = 6, i.e. 63
For b = 4, a = 7, i.e. 74
For b = 5, a = 8, i.e. 85
For b = 6, a = 9, i.e. 96
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संबंधित प्रश्न
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3 A
+ 2 5
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B 2
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Find the values of the letters in the following and give reasons for the steps involved
A B
x 3
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C A B
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Find the values of the letters in the following and give reasons for the steps involved.
A B
x 5
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C A B
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Find the values of the letters in the following and give reasons for the steps involved.
A B
x 6
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B B B
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Find the values of the letters in the following and give reasons for the steps involved.
1 2 A
+ 6 A B
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A 0 9
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| A | |
| + | A |
| + | A |
| B | A |
Find the value of the letters in the following question.
C B A
+ C B A
1 A 3 0
Find the value of the letters in the following question.
A 0 1 B
+ 1 0 A B
B 1 0 8
Find the value of the letters in the following question.
A B
× 6
C 6 8
Work out the following multiplication.
12345679
× 9
Use the result to answer the following question.
What will be 12345679 × 63?
