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Question Number 113003 by Aina Samuel Temidayo last updated on 10/Sep/20

The digits of a three−digit number A are written in the reverse order to form another three−digit number B. If B>A and B−A is perfectly divisible by 7. Find the range of values of A.

$$\mathrm{The}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{a}\:\mathrm{three}−\mathrm{digit}\:\mathrm{number}\:\mathrm{A} \\ $$ $$\mathrm{are}\:\mathrm{written}\:\mathrm{in}\:\mathrm{the}\:\mathrm{reverse}\:\mathrm{order}\:\mathrm{to} \\ $$ $$\mathrm{form}\:\mathrm{another}\:\mathrm{three}−\mathrm{digit}\:\mathrm{number}\:\mathrm{B}. \\ $$ $$\mathrm{If}\:\mathrm{B}>\mathrm{A}\:\mathrm{and}\:\mathrm{B}−\mathrm{A}\:\mathrm{is}\:\mathrm{perfectly} \\ $$ $$\mathrm{divisible}\:\mathrm{by}\:\mathrm{7}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of} \\ $$ $$\mathrm{values}\:\mathrm{of}\:\mathrm{A}. \\ $$

Commented byRasheed.Sindhi last updated on 10/Sep/20

miss^(?) /mr^(?) Aina pl read my comment (Is the answer 8?) of Q#112810

$$\overset{?} {{miss}}/\overset{?} {{mr}}\:{Aina}\:{pl}\:{read}\:{my}\:{comment} \\ $$ $$\left({Is}\:{the}\:{answer}\:\mathrm{8}?\right)\:{of}\:{Q}#\mathrm{112810} \\ $$

Commented byAina Samuel Temidayo last updated on 10/Sep/20

Replied✓

$$\mathrm{Replied}\checkmark \\ $$

Answered by Rasheed.Sindhi last updated on 14/Sep/20

A=100a+10b+c B=100c+10b+a Given that:(1)B>A⇒c>a (2)7∣B−A 100(c−a)+a−c 100(c−a)−(c−a) 99(c−a) 7 ∣ 99(c−a) ∵ 7∈P ∧ 7 ∤ 99 ∴ 7∣ c−a ∴ c−a=7k c=a+7k As c>a so k≠0 And for a,c are decimal digits of a 3-digit number so 1≤a,c≤9 ∴ a=1,2 & k=1⇒c=8,9 (a,c)=(1,8) ∧ (a,c)=(2,9) A=100+10b+8 , 200+10b+9 {: ((A=108,118,128,...,198)),(( AND)),((A=209,219,229,...,299)) }Ans

$${A}=\mathrm{100}{a}+\mathrm{10}{b}+{c} \\ $$ $${B}=\mathrm{100}{c}+\mathrm{10}{b}+{a} \\ $$ $${Given}\:{that}:\left(\mathrm{1}\right){B}>{A}\Rightarrow{c}>{a} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right)\mathrm{7}\mid{B}−{A} \\ $$ $$\mathrm{100}\left({c}−{a}\right)+{a}−{c} \\ $$ $$\mathrm{100}\left({c}−{a}\right)−\left({c}−{a}\right) \\ $$ $$\mathrm{99}\left({c}−{a}\right) \\ $$ $$\mathrm{7}\:\mid\:\mathrm{99}\left({c}−{a}\right) \\ $$ $$\because\:\mathrm{7}\in\mathbb{P}\:\wedge\:\mathrm{7}\:\nmid\:\mathrm{99} \\ $$ $$\therefore\:\mathrm{7}\mid\:{c}−{a} \\ $$ $$\therefore\:{c}−{a}=\mathrm{7}{k} \\ $$ $$\:\:\:\:\:{c}={a}+\mathrm{7}{k} \\ $$ $${As}\:{c}>{a}\:{so}\:{k}\neq\mathrm{0} \\ $$ $${And}\:{for}\:{a},{c}\:{are}\:{decimal}\:{digits} \\ $$ $${of}\:{a}\:\mathrm{3}-{digit}\:\:{number} \\ $$ $${so}\:\mathrm{1}\leqslant{a},{c}\leqslant\mathrm{9} \\ $$ $$\therefore\:{a}=\mathrm{1},\mathrm{2}\:\&\:{k}=\mathrm{1}\Rightarrow{c}=\mathrm{8},\mathrm{9} \\ $$ $$\left({a},{c}\right)=\left(\mathrm{1},\mathrm{8}\right)\:\wedge\:\left({a},{c}\right)=\left(\mathrm{2},\mathrm{9}\right) \\ $$ $${A}=\mathrm{100}+\mathrm{10}{b}+\mathrm{8}\:,\:\mathrm{200}+\mathrm{10}{b}+\mathrm{9} \\ $$ $$ \\ $$ $$\left.\begin{matrix}{{A}=\mathrm{108},\mathrm{118},\mathrm{128},...,\mathrm{198}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{AND}}\\{{A}=\mathrm{209},\mathrm{219},\mathrm{229},...,\mathrm{299}}\end{matrix}\right\}\mathrm{Ans} \\ $$

Commented byAina Samuel Temidayo last updated on 11/Sep/20

Thanks.

$$\mathrm{Thanks}. \\ $$

Commented byRasheed.Sindhi last updated on 12/Sep/20

The last part of my answer is erroneous. I′ll correct it.

$$\mathcal{T}{he}\:{last}\:{part}\:{of}\:{my}\:{answer}\:{is}\: \\ $$ $${erroneous}.\:{I}'{ll}\:{correct}\:{it}. \\ $$

Commented byRasheed.Sindhi last updated on 14/Sep/20

Corrected.

$${Corrected}. \\ $$

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