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The digits of a positive integer having three digits are in AP and sum of their digits is 21. The number obtained by reversing the digits is 396 less than the original number. Find the original number.
- a)876
- b)579
- c)975
- d)678
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The digits of a positive integer having three digits are in AP and sum...
Let the digits at ones, tens and hundreds place be (a−d)a and (a+d) respectively. The, the number is
(a+d)×100+a×10+(a−d) = 111a+99d
The number obtained by reversing the digits is
(a−d)×+a×10+(a+d) = 111a−99d
It is given that the sum of the digits is 21.
(a−d)+a+(a+d) = 21 ...(i)
Also it is given that the number obtained by reversing the digits is 594 less than the original number.
∴111a−99d = 111a+99d−396 ...(ii)
⟹ 3a = 21 and 198d = 396
⟹ a = 7 and d = -2
Original number = (a−d)×+a×10+(a+d)
= 100(9) + 10(7) + 5
= 975
(a+d)×100+a×10+(a−d) = 111a+99d
The number obtained by reversing the digits is
(a−d)×+a×10+(a+d) = 111a−99d
It is given that the sum of the digits is 21.
(a−d)+a+(a+d) = 21 ...(i)
Also it is given that the number obtained by reversing the digits is 594 less than the original number.
∴111a−99d = 111a+99d−396 ...(ii)
⟹ 3a = 21 and 198d = 396
⟹ a = 7 and d = -2
Original number = (a−d)×+a×10+(a+d)
= 100(9) + 10(7) + 5
= 975
Most Upvoted Answer
The digits of a positive integer having three digits are in AP and sum...
If u can prepare for the jee then u can use this pattern
three numbers digit is such that - abc
a+b+c=21 a,b,c in a.p. so - 2b=a+c then 3b=21 b=7
abc-369=cbaafter this u can check the option 975-369=579this is only option who confirms this option
bs plzz yar ans psnd aaye to like kr dena or ek reply kr dena ok
three numbers digit is such that - abc
a+b+c=21 a,b,c in a.p. so - 2b=a+c then 3b=21 b=7
abc-369=cbaafter this u can check the option 975-369=579this is only option who confirms this option
bs plzz yar ans psnd aaye to like kr dena or ek reply kr dena ok
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The digits of a positive integer having three digits are in AP and sum...
Let the three digits of the positive integer are a-d,a and a+d sum of digits=21 a-d+a+a+d=21 3a=21 a=7 original no=100(a-d)+10a+a+d=100(7-d)+10(7)+7+d=700+70+7-99d=777-99d number formed by reversing the digits=100(a+d)+10a+a-d=100(7+d)+10(7)+7-d=777+99d given 777-99d=396+777+99d -99d-99d=396 -198d=396 d=-2 a-d=7-(-2)=9 a=7 a+d=7+(-2)=5 original no=100(9)+10(7)+5=975
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The digits of a positive integer having three digits are in AP and sum of their digits is 21. The number obtained by reversing the digits is 396 less than the original number. Find the original number.a)876b)579c)975d)678Correct answer is option 'C'. Can you explain this answer?
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