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Number Theory Level 3

How many ordered sets of digits $(a, b)$ are there, such that the number $\overline {1ab2}$ is a multiple of 3?

Details and assumptions

Digits are integers from 0 to 9 inclusive.

The notation $\overline{abc}$ denotes $100a + 10b + 1c$ , as opposed to $a \times b \times c$ . As an explicit example, for $a=2, b=3, c=4$ , $\overline{abc} = 234$ and not $2 \times 3 \times 4 = 24$ .

The answer is 34.

6 solutions

Muhammad Shariq
Dec 2, 2013

$\overline{1ab2}$ is divisible by 3 if the sum of its digits is divisible by 3, i.e $3|3+a+b$ . Since $3|1+2$ , in order for $3$ to divide $3+a+b$ , 3 must divide $a + b$ . We also know that $a$ and $b$ are integers such that $0 \le a \le 9$ and $0 \le b \le 9$ which implies that $0 \le a +b \le 18$ . Since $3|a+b$ , we have that $a+b=0,3,6,9,12,15,18$ . We now consider each case separately and find the number of ordered pairs $(a,b)$ .

Case 1: When $a+b=0$ , we get $1$ ordered pair; $(a,b)=(0,0)$ .

Case 2: When $a+b=3$ , we get $4$ ordered pairs; $(a,b)=(3,0),(0,3),(2,1),(1,2)$ .

Case 3: When $a+b=6$ , we get $7$ ordered pairs; $(a,b)=(6,0),(0,6),(1,5),(5,1),(4,2),(2,4),(3,3)$ .

Case 4: When $a+b=9$ , we get $10$ ordered pairs; $(a,b)=(9,0),(0,9),(1,8),(8,1),(7,2),(2,7),(3,6),(6,3),(4,5),(5,4)$ .

Case 5: When $a+b=12$ , we get $7$ ordered pairs; $(a,b)=(3,9),(9,3),(4,8),(8,4),(5,7),(7,5),(6,6)$ .

Case 6: When $a+b=15$ , we get $4$ ordered pairs; $(a,b)=(6,9),(9,6),(7,8),(8,7)$ .

Case 7: When $a+b=18$ , we get $1$ ordered pairs; $(a,b)=(9,9)$ .

Hence the number of ordered pairs $(a,b)$ is $1+4+7+10+7+4+1=\boxed{34}$ .

In case 7, it should've been a + b = 18 instead of a + b = 6. Sorry about that.

Muhammad Shariq - 7 years, 6 months ago

I've made the edit. In future, remember to check that your mathematical statements display correctly.

Calvin Lin Staff - 7 years, 6 months ago

$A\quad linear\quad equation\quad :\quad { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4+.......+ }{ x }_{ k-1 }+{ x }_{ k }=m\\ have\quad \quad \quad \quad { C }(m-1,k-1)\quad \quad solutions,\quad when\quad \overset { k }{ \underset { i=1 }{ { x }_{ i } } } >0\\ and\quad \quad \quad { C }(m+k-1,k-1)\quad solutions,\quad when\quad \overset { k }{ \underset { i=1 }{ { x }_{ i } } } \ge 0\\ but\quad I\quad just\quad forgot\quad the\quad restriction\quad 0\le a,b\le 9\quad .\\ So,\quad whenever\quad a+b=\quad 12,15,18\\ we\quad cannot\quad just\quad follow\quad the\quad general\quad conclusions\quad directly.\\ To\quad find\quad the\quad ordered\quad solution\quad of\quad a+b\quad =\quad 12\quad \\ is\quad just\quad finding\quad the\quad number\quad of\quad ordered\quad solutions\quad \\ of\quad a+b=(18-12)=6.\\ [for\quad a+b>9\quad we\quad replace\quad a\quad \& \quad b\quad with\quad \\ { x }_{ 1 }\quad \& \quad { x }_{ 2 }\quad :{ \quad x }_{ 1 }=9-a\quad \& \quad { x }_{ 2 }=9-b\\ \therefore \quad for\quad a+b>9\quad the\quad eqn.\quad will\quad be\\ \quad { x }_{ 1 }+{ x }_{ 2 }=18-(a+b)\quad ;\quad { x }_{ 1 },{ x }_{ 2 }\ge 0]\\ \\ Now,\quad number\quad of\quad ordered\quad solutions\quad of\quad \\ a+b=0\quad \equiv \quad a+b=18\quad \\ a+b=3\quad \equiv \quad a+b=15\\ a+b=6\quad \equiv \quad a+b=12\\ [a+b=9\quad \equiv \quad a+b=9]\\ \\ **NOTE:(here\quad number\quad of\quad ordered\quad solutions\quad \\ are\quad equal,not\quad the\quad solutions\quad themselves)\\ \\ Now\quad you\quad have\quad already\quad discuss\quad the\quad cases.\quad \\ \therefore \quad a+b=6\quad have\quad C(6+2-1,2-1)=7\quad ordered\quad solutions\quad \\ so\quad a+b=12\quad have\quad 7\quad no.\quad of\quad ordered\quad solution\\ and\quad so\quad on...$

Subhajit Ghosh - 6 years, 6 months ago

Arkan Megraoui
Dec 4, 2013

The divisibility rule for 3 tells us that "A number is divisible by 3 if, and only if, its digits sum is divisible by 3".

This means that we want $1+a+b+2\equiv 0\pmod 3\Leftrightarrow a+b\equiv 0\pmod 3$ .

This is only possible if

$a\equiv b\equiv 0\pmod 3$ or

$(a,b)\equiv (1,2)\pmod 3$ or

$(a,b)\equiv (2,1)\pmod 3$ .

The first case is possible when $a,b$ are congruent to one of $\{0,3,6,9\}$ , which is possible in $4^2=16$ ways.

The last 2 cases are possible when $a,b$ are congruent to one of $1,4,7$ and $2,5,8$ in some order, for a total of $2\cdot 3^2=18$ ways.

So the total number of ways is $16+18=\boxed{34}$ .

Dishant Shah
Dec 3, 2013

a+b=multiple of 3 (since to be divisible by 3 the sum of the digits of the no.(here 1+a+b+2) should be divisible by 3 )hence ab can be 03,06,09,12,15,18,........30,33,36,39,.........90,93,96,99. Hence we can observe that there are 3 possibilities in each set of ten except in the case of a = 3,6,9. however we overlook the fact that even ab=00 satisfies the required condition. hence in total there are 34 solutions

Narasimha Murthy
Mar 5, 2014

Solution: 34 Ordered Pairs

Key aspect of the solution:

For the number 1ab2 to be multiple of 3, the sum of the digits 1 + a + b + 2 must be divisible by 3.

We already have:

With the two extreme digits summing to 3, we are left with finding all the ordered pairs (a,b) that are multiples of 3.

It is sufficient to find the ordered pairs as per the condition below to form the full set of solutions:

The set of numbers that fit the ordered pair (a,b) being multiple of 3 is {0, 3, 6, 9, 12, 15, 18}.

Elaboration of ordered pairs for each element of the solution set:

1 ordered pair for element "0"; (0,0) is the only ordered pair that satisfies this element.

4 ordered pairs for element "3"; (0,3), (1,2), (2,1), (3,0) are the ordered pairs for this element.

7 ordered pairs for element "6"; (0,6), (1,5), (2,4), (3,3), (4,2), (5,1), (6,0) are the ordered pairs for this element.

10 ordered pairs for element "9"; (0,9), (1,8), (2,7), ..., (8,1), (9,0) are the ordered pairs for this element.

7 ordered pairs for element "12"; (3,9), (4,8), ..., (8,4), (9,3) are the ordered pairs for this element.

4 ordered pairs for element "15"; (6,9), (7,8), (8,7), (9,6) are the ordered pairs for this element.

1 ordered pair for the element "18"; (9,9) is the only ordered pair that satisfies this element.

Anthony J.
Dec 8, 2013

We can proceed with casework on the sum of $a+b$ , as the sum of the digits of any number divisible by 3 is divisible by 3. Thus, the sum $a+b$ can equal 0, 3,6,9,12,15, or 18.

*if a+b=0 * Then there is $1$ are legal ordered pair

if a+b=3 Then there are $4$ are legal ordered pairs

*if a+b=6 * Then there are $7$ are legal ordered pairs

*if a+b=9 * Then there are $10$ are legal ordered pairs

*if a+b=12 * Then there are $7$ are legal ordered pairs

*if a+b=15 * Then there are $4$ are legal ordered pairs

*if a+b=18 * Then there is $1$ are legal ordered pair

The total is thus $1+4+7+10+7+4+1= \boxed{34}$ legal ordered pairs

Muralidhar Kamidi
Dec 4, 2013

Since 1 and 2 already add up to 3, the sum of a and b must be a multiple of 3. Here are the possibilities: 3: 12, 21, 03, 30

6: 33, 42, 24, 15, 51, 06, 60

9: 18, 81, 72, 27, 63, 36, 09, 90, 45, 54

12: 39, 93, 84, 48, 57, 75, 66

15: 96, 69, 78, 87

18: 99

0: 00

That is 34 of them in all. I know there would be an easier way.

1, 2 buckle my shoe

The answer is 34.

6 solutions

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