How many 5 digit positive integers are there such that the unit's place is 4 and it is divisible by 6?

Nice proofs and solutions are **always** welcome!

How many 5 digit positive integers are there such that the unit's place is 4 and it is divisible by 6?

Nice proofs and solutions are **always** welcome!

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TopNewestFill the thousands, hundreds, and tens digits arbitrarily. Now there are exactly three ways to fill the ten-thousands digit. This easily gives \(10^3 \cdot 3 = \boxed{3000}\) numbers. – Ivan Koswara · 2 years ago

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There are 15000 5 digit numbers that are divisible by 6.The first i.e the smallest one ending in 2.The unit digit of these numbers make a pattern,2,8,4,0,6,2,8,4,0,6,2....Hence 1 in every 5 consecutive numbers ends in 4.Hence answer to the problem is ,15,000/5=3000. – Adarsh Kumar · 2 years ago

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– Kalash Verma · 2 years ago

Nice solution!Log in to reply

– Adarsh Kumar · 2 years ago

Thanx bud!Log in to reply

– Adarsh Kumar · 2 years ago

Damn simple prob!Two second question!Log in to reply

The numbers are of the form 10k+4. We set 10k+4=6z or 5k+2=3z. We see that when k=-1 and z = -1 there is a solution note that when k= -1+3b and z = -1+5b. 5k+2= 5(-1+3b)+2=15b-3=3(5b-1)=3z. By euclidean algorithm there is only one solution in the interval where k= -2,-1,0 and z= -4,-3,-2,-1,0. eg one solution for (k,w) in this specified domain. Now we have 6z=6(-1+5b)=30b-6. All such numbers are of this form. Now 10000=<30b-6<=99999. 334<=b<=3333. Number of integers in the range = no of integers from 1 to 3333 - no of integers from 1 to 333 = 3333-333=3000 – Siddharth Iyer · 1 year, 11 months ago

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First term is 10014 and last term is 99984 (can easily be found by hit and trial method). These are the corresponding terms of an arithmetic progression with common difference = 30. An = a + (n-1)

d, 99984 = 110014 + (n-1)30, n = 3000 – Rahul Mehta · 2 years agoLog in to reply

30 – Shayaan Subzwari · 2 years ago

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It can also be seen as the number of four digit numbers that are divisible by 3, because the rest of the digits simply need to add upto a number that is one less than a multiple of 3. Under 1000 you have 333 such numbers and under a 10000 you have 3333 numbers divisible by 3. So we get 3000 by subtraction. – Vishnu C · 2 years ago

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Please help me in my approach:

Let the required \(5\) digit number be \(\overline{a_1a_2a_3a_44}\)

Since the number is already divisible by \(2\) we need to assure the divisibility of \(3\)

So , \(\displaystyle\sum_{i = 1}^4 a_i \equiv -4 \equiv 2 \pmod{3}\)

So , \(\displaystyle\sum_{i =1}^4 a_i = 3k+2\) where \(\begin{cases} 1 \leq k \leq 11 \\ 1 \leq a_1 \leq 9 \\ 0 \leq a_2,a_3,a_4 \leq 9 \end{cases}\)

Indeed , finding solutions to this equation will fetch the required answer. – Nihar Mahajan · 2 years ago

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– Adarsh Kumar · 2 years ago

No need to complicate it this much bud!Log in to reply

– Nihar Mahajan · 2 years ago

But we must learn different methods.May be this method can help in some other problem.Log in to reply

– Parth Lohomi · 2 years ago

Learn mthods that increase problem solving speed and increase accuracy, complicating things may take you down. :)Log in to reply

– Nihar Mahajan · 2 years ago

Of course , a short and quick method is always preferred.But we must know different approaches to solve.May be the "short and quick" method does not work in some problems.In that case such different approaches may prove useful.Hence I prefer doing the same problem using different methods , rather that leaving it by solving it with the shortest method. :)Log in to reply

– Adarsh Kumar · 2 years ago

But if there are two different methods to solve a prob we must take the shorter solution.Donn'y u think?Log in to reply

– Nihar Mahajan · 2 years ago

I know we must have the shorter one , but if the shorter method is not applicable to some other problem then what will you do? If we do not get a pattern in some other problem this method will not work.Log in to reply

– Adarsh Kumar · 2 years ago

True!Let me tell you the method of solving this.Because \(a_i\leq 9\),you can assume \(a_1=9-x,a_2=9-y,a_3=9-z,a_4=9-p\).Notice here that \(x,y,z,p\) are non-negative integers.Now you can consider every case and use stars and bars.But it would be very tedious.Log in to reply

– Nihar Mahajan · 2 years ago

Yeah , thats a great idea. Let me see what happens next.Log in to reply

we form a AP here : 10014, 10044, 10074............................................................99984 therefore no. of terms n= 99984-10014/30 +1 =3000 – Aakash Khandelwal · 2 years ago

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