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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-

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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.
$$\mathrm{Two}\:\mathrm{different}\:\mathrm{prime}\:\mathrm{numbers}\:\mathrm{between} \\ $$$$\mathrm{4}\:\mathrm{and}\:\mathrm{18}\:\mathrm{are}\:\mathrm{chosen}.\:\mathrm{When}\:\mathrm{their}\:\mathrm{sum}\:\mathrm{is} \\ $$$$\mathrm{subtracted}\:\mathrm{from}\:\mathrm{their}\:\mathrm{product}\:\mathrm{then}\:\mathrm{a} \\ $$$$\mathrm{number}\:{x}\:\mathrm{is}\:\mathrm{obtained}\:\mathrm{which}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{multiple}\:\mathrm{of}\:\mathrm{17}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{digits}\:\mathrm{of} \\ $$$$\mathrm{number}\:{x}. \\ $$
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 .
$$\mathrm{let}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:. \\ $$$$\:\:\:\:\:\mathrm{pq}−\left(\mathrm{p}+\mathrm{q}\right)=\mathrm{17n} \\ $$$$\mathrm{p},\:\mathrm{q}\:\mathrm{can}\:\mathrm{be}\:\:\mathrm{among}\:\:\mathrm{5},\mathrm{7},\mathrm{11},\mathrm{13},\mathrm{17} \\ $$$$\mathrm{if}\:\mathrm{p}=\mathrm{17},\:\mathrm{q}=\mathrm{13} \\ $$$$\mathrm{15}^{\mathrm{2}} −\mathrm{4}−\mathrm{30}=\mathrm{191}\neq\mathrm{17n} \\ $$$$\mathrm{if}\:\mathrm{p}=\mathrm{17},\:\mathrm{q}=\mathrm{11} \\ $$$$\mathrm{196}−\mathrm{9}−\mathrm{28}=\mathrm{159}\neq\mathrm{17n} \\ $$$$\mathrm{if}\:\mathrm{p}=\mathrm{17},\:\mathrm{q}=\mathrm{7} \\ $$$$\mathrm{144}−\mathrm{25}−\mathrm{24}=\mathrm{95}\neq\mathrm{17n} \\ $$$$\mathrm{if}\:\mathrm{p}=\mathrm{17},\:\mathrm{q}=\mathrm{5} \\ $$$$\mathrm{85}−\mathrm{22}=\mathrm{63}\neq\mathrm{17n} \\ $$$$…………………………………. \\ $$$$\mathrm{if}\:\mathrm{p}=\mathrm{13},\:\mathrm{q}=\mathrm{11} \\ $$$$\mathrm{143}−\mathrm{24}=\mathrm{119}=\mathrm{17}×\mathrm{7} \\ $$$$\mathrm{if}\:\mathrm{p}=\mathrm{13},\:\mathrm{q}=\mathrm{7} \\ $$$$\mathrm{91}−\mathrm{20}=\mathrm{71}\neq\mathrm{17n} \\ $$$$\mathrm{if}\:\mathrm{p}=\mathrm{13},\:\mathrm{q}=\mathrm{5} \\ $$$$\mathrm{65}−\mathrm{18}=\mathrm{47}\neq\mathrm{17n} \\ $$$$\mathrm{if}\:\mathrm{p}=\mathrm{11},\:\mathrm{q}=\mathrm{7} \\ $$$$\mathrm{77}−\mathrm{18}=\mathrm{59}\neq\mathrm{17n} \\ $$$$\mathrm{if}\:\mathrm{p}=\mathrm{11},\:\mathrm{q}=\mathrm{5} \\ $$$$\mathrm{55}−\mathrm{16}=\mathrm{39}\neq\mathrm{17n} \\ $$$$\mathrm{if}\:\mathrm{p}=\mathrm{7},\:\mathrm{q}=\mathrm{5} \\ $$$$\mathrm{35}−\mathrm{12}=\mathrm{23}\neq\mathrm{17n} \\ $$$$\mathrm{Hence}\:\mathrm{x}=\mathrm{17}×\mathrm{7}=\mathrm{119} \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{its}\:\mathrm{digits}=\mathrm{11}\:. \\ $$
Commented by Tinkutara last updated on 13/Aug/17
So totally hit and trial method is applicable?
$$\mathrm{So}\:\mathrm{totally}\:\mathrm{hit}\:\mathrm{and}\:\mathrm{trial}\:\mathrm{method}\:\mathrm{is} \\ $$$$\mathrm{applicable}? \\ $$
Commented by ajfour last updated on 13/Aug/17
dint know a decent way out..
$$\mathrm{dint}\:\mathrm{know}\:\mathrm{a}\:\mathrm{decent}\:\mathrm{way}\:\mathrm{out}.. \\ $$

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