The digits of a three digit number N are in A.P.If sum of the digits is $$15$$ and the number obtained by $$re-versing$$ the digits of the number is $$594$$ less than the original number, then $$1000N$$−$$252$$ is equal to

Question

The digits of a three digit number N are in A.P.If sum of the digits is $$15$$ and the number obtained by $$re-versing$$ the digits of the number is $$594$$ less than the original number, then $$1000N$$−$$252$$ is equal to

Infinity Learn · Accepted Answer

Let three digits of N be: Hundred’s digit $$=$$ a – dten’s digit $$=$$ a,and unit’s digit $$=$$ $$a+d(a$$–$$d)+a+(a+d)=15$$ fi $$a=5andN=100(5$$−$$d)+50+(5+d)$$                                         (1)Number$$ obtained by reversing digits of N $$isN1=100(5+d)+50+(5$$−$$d)                                       $$(2)Now$$,          $$594=N$$−$$N1=100($$−$$2d)+2d$$⇒    $$2d=$$−$$6$$⇒$$d=$$−$$3Thus$$, N $$=$$ $$852$$ $$and1000N$$−$$252=1000600=53$$

The digits of a three digit number $N$ are in A.P.If sum of the digits is 15 and the number obtained by re-versing the digits of the number is 594 less than the original number, then $\frac{1000}{N - 252}$ is equal to

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The digits of a three digit number N are in A.P.If sum of the digits is 15 and the number obtained by re-versing the digits of the number is 594 less than the original number, then 1000N−252 is equal to

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The digits of a three digit number $N$ are in A.P.If sum of the digits is 15 and the number obtained by re-versing the digits of the number is 594 less than the original number, then $\frac{1000}{N - 252}$ is equal to