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Question Number 19631 by Tinkutara last updated on 13/Aug/17

Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product then a number x is obtained which is a multiple of 17. Find the sum of digits of number x.

Twodifferentprimenumbersbetween4and18arechosen.Whentheirsumissubtractedfromtheirproductthenanumberxisobtainedwhichisamultipleof17.Findthesumofdigitsofnumberx.

Answered by ajfour last updated on 13/Aug/17

let p and q . pq−(p+q)=17n p, q can be among 5,7,11,13,17 if p=17, q=13 15^2 −4−30=191≠17n if p=17, q=11 196−9−28=159≠17n if p=17, q=7 144−25−24=95≠17n if p=17, q=5 85−22=63≠17n ........................................ if p=13, q=11 143−24=119=17×7 if p=13, q=7 91−20=71≠17n if p=13, q=5 65−18=47≠17n if p=11, q=7 77−18=59≠17n if p=11, q=5 55−16=39≠17n if p=7, q=5 35−12=23≠17n Hence x=17×7=119 Sum of its digits=11 .

letpandq.pq−(p+q)=17np,qcanbeamong5,7,11,13,17ifp=17,q=13152−4−30=191≠17nifp=17,q=11196−9−28=159≠17nifp=17,q=7144−25−24=95≠17nifp=17,q=585−22=63≠17n........................................ifp=13,q=11143−24=119=17×7ifp=13,q=791−20=71≠17nifp=13,q=565−18=47≠17nifp=11,q=777−18=59≠17nifp=11,q=555−16=39≠17nifp=7,q=535−12=23≠17nHencex=17×7=119Sumofitsdigits=11.

Commented by Tinkutara last updated on 13/Aug/17

So totally hit and trial method is applicable?

Sototallyhitandtrialmethodisapplicable?

Commented by ajfour last updated on 13/Aug/17

dint know a decent way out..

dintknowadecentwayout..

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