Question

A maths detective is in search of a five-digit number. He knows that its last digit is 3 times the first digit, the 3rd digit is 3 less than the second digit, and the 4th digit is 4 more than the third digit. If the number is divisible by 3, what will the 1st digit of the number be?

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

The correct option is C 2

$\text{Expressing the first two digits as variables,}\text{we get:}$

$1stdigit=a$
$2nddigit=b$

$\text{Now, the rest 3 digits are related to}a\text{and}b\text{as follows:}$

$3rddigit=3lessthanthe2nddigit=b–3$

$4thdigit=4morethanthe3rddigit=(b–3)+4=b+1$

$5thdigit=3timesthe1stdigit=3a$

$Sumofthedigits=a+b+(b–3)+(b+1)+3a=4a+3b–2$

$\text{Now, we know that the five-digit number}\text{is a multiple of 3.}$

$⟹4a+3b–2=3k.........\left(forsomenaturalnumberk\right)$
$⟹4a–2=3(k–b)$
$⟹2(2a–1)=3(k–b)$

$\text{Hence,}2(2a–1)\text{is a multiple of 3.}$
$⟹(2a–1)\text{is a multiple of 3.}$

$\text{A digit can only take values from 0 to 9.}\text{Also, the 5th digit is}3a.$

$\text{Hence, the only possible values for the}\text{5th digit are 3, 6, or 9.}$

$Case1:5thdigit=3$
$⟹3a=3⟹a=1$
$⟹(2a–1)=2\times 1–1=1$
$\text{1 is}not\text{a multiple of 3.}$
$\therefore a\ne 1$

$Case2:5thdigit=6$
$⟹3a=6⟹a=2$
$⟹(2a–1)=2\times 2–1=3$
$\text{3}is\text{a multiple of 3.}$
$\therefore a=2$

$Case3:5thdigit=9$
$⟹3a=9⟹a=3$
$⟹(2a–1)=2\times 3–1=5$
$\text{5 is}not\text{a multiple of 3.}$
$\therefore a\ne 3$

Hence, the 1st digit of the five-digit number is 2.