A Chest Of Gold To Divide Amongst Friends

Let the number of gold coins be $n$ .Then,from the problem we have:

$n=6k+4$

$n=5p+3$
Doing just a litttle manipuation on both equations we get:

$n=6(k+1)-2$

$n=5(p+1)-2$

Logically, we realise that $6(k+1)=5(p+1)$ . These numbers are divisible by $6$ and $5$ simultaneously.Therefore they are also divisible by $6*5=30$ .So,these numbers are multiple of $30$ .Since we want the minimum, the smallest possible result is $30$ .Therefore $6(k+1)=30$ where we get $k=4$ and also $5(p+1)=30$ where we get $p=5$ .From this we get $n=\boxed{28}$ .When divided by $7$ this number leaves a remainder $0$ or $28=7*4+0$ thus , the remainder is $\boxed{0}$ .

1. List out the numbers that have remainder 4 when divided by 6: 4, 10, 16, 22, 28 , 34, 40, ...
2. List out the numbers that have remainder 3 when divided by 5: 3, 8, 13, 18, 23, 28 , 33, ...
3. Find out the common number between them: 28
4. Divide that by 7 and find the remainder: $\boxed0$

5k+3=6n+4

5k-6n=1

Obvious k=5,n=4

Total coins=28

28%7=0

Let the total number of gold coins be ₌ x

x ₌ 6a + 4 …..(1)

x ₌ 5b + 3 …..(2)

Equating the two gives,

6a + 4 ₌ 5b + 3

i.e 6a – 5b ₌ -1

Testing integers on the equation, the pair that gives the smallest number of gold coins in the box is 4 and 5 for a and b respectively. Thus,

x ₌ 6(4) + 4 ₌ 28.

x ₌ 5(5) + 3 ₌ 28

Therefore 28/7 ₌ 4 remainder 0 as required.

Let the no. of coins be $x$ . Let $p$ and $q$ be arbitrary constants. From the question, we get----

$x=6p+4$ ....(i) and $x=5q+3$ ....(ii)

From (i) and (ii), we get----

$6p+4=5q+3 \implies 6p-5q+1=0$

The equation is satisfied for the least positive values of $p,q$ with $p=4$ and $q=5$ . So, we get the no. of coins as $x=6p+4=6\times 4 + 4 = 24+4 = 28$

Now, if 28 coins are equally distributed among 7 people, then each will get 4 coins and 0 coins are left over. So, the answer is $\boxed{0}$

List a few multiples of 5 and 6 . Add 3 and 4 to them respectively. you will find that the number 28 is common in both lists( list after adding 3 and 4). When 28 is divided by 7, the remainder is zero . Therefore, answer is 0

Be $c$ the number of coins. Then $c \equiv 4 \mod 6 , c \equiv 3 \mod 5$ . It has got solutions for the Chinese remainder theorem , the smallest of which is $28 \equiv \boxed{0} \mod 7$ .

We are going to proceed using The Chinese Remainder Theorem.

Let $x$ be the number of coins contained in the box, we have:

• $x\equiv 4(mod\ 6)$
• $x\equiv 3(mod\ 5)$ *

First, we have $x\equiv 4(mod\ 6)\ \Rightarrow \ x=6k+4$ ** with $k$ an integer.

Substituting this result into * gives: $6k+4\equiv 3(mod\ 5)$ hence $6k\equiv -1(mod\ 5)$ .

Now, what we should try to do here is to get rid of that $6$ , to have an expression which only contains $k$ . To do that, we should find the modular multiplicative inverse of $6$ modulo $5$ . We know that $6\equiv 1(mod\ 5)$ which means that $1$ is the modular multiplicative inverse of $6$ modulo $5$ .

In cases of bigger numbers, one should use the euclidean algorithm to determine the modular multiplicative inverse.

Hence $6\times 1\times k\equiv -1\times 1(mod\ 5)\Rightarrow k\equiv -1(mod\ 5)$ , which means $k$ can be written as: $k=5i-1$ with $i$ an integer.

Substituting this result into ** gives: $x=6(5i-4)+4$ which means $x=30i-2$

Since $x>0$ , the minimum value $x$ can take is $\boxed{28}$ , attained with $i=1$

So the minimum number of coins that satisfies the two desired conditions is $28$ .

And we know that $28\equiv 0(mod\ 7)$ , which means that, if we equally divide these coins among 7 people, no coins will be left.

Hence, the result is $\boxed{0}$ .

x = 4 (mod 6) => a1 (mod m1)

x = 3 (mod 5) => a2 (mod m2)

by the Chinese remainder theorem , we know there exists a solution in the form :

x = a1 * m2 *(the multiplicative inverse of m2) + a2 * m1 * (the multiplicative inverse of m1)

the multiplicative inverse of 5 (mod 6) = 5 and the multiplicative inverse of 6 (mod 5) = 1

x = 4 * 5 * 4 + 3 * 6 * 1 = 118 = 28 (mod 30) which is the smallest amount of coins that meet the conditions, the solution then is obvious there is no coin left over

There is an easy way. You know that there is 3 coins left after divided by 5. Therefore number has to end with either 3 or 8 due to the fact that multiples of 5 end with 0 and 5. After that you go from 13 to 18 to 23 to 28 and oh that works!

Let N be the number of gold coins where

• $N \pmod{6} \equiv 4$
• $N \pmod{5} \equiv 3$

By trial-and-error, If $N = 6$ , $6 \pmod{6} \equiv 0$

So, I add 4 to 6 to make $N$ follow the first premise. If $N = 10$ , $10 \pmod{6} \equiv 4$ and $10 \pmod{5} \equiv 0$

Continue adding 6 to $N$ until you find the value of $N$ that will follow both premises. If $N = 16$ , $16 \pmod{6} \equiv 4$ and $16 \pmod{6} \equiv 1$

If $N = 22$ , $22 \pmod{6} \equiv 4$ and $22 \pmod{5} \equiv 2$

If $N = 28$ , $28 \pmod{6} \equiv 4$ and $28 \pmod{5} \equiv 3$

Therefore, there are 28 gold coins.

When 28 coins are to be equally divided among 7 people, there are $28 \pmod{7} \equiv 0$ gold coins left over.

You can easily apply the Chinese Remainder Theorem. Set up the two modular functions $x= 3 \mod (5)$ and $x=4\mod(6)$ .

Next, let $x=6j+4$ .

Thus, $6j+4=3\mod(5)$ . From here, solve for $j$ .

$j=4\mod(5)$ .

Then, let $j=5k+4$ , as $5k+4= 4\mod(5)$ is clearly true.

Substitute $j$ with $5k+4$ in $x=6j+4$ , so $x=30k+28$ .

That means that $x=28$ . You can check that this is true easily.

$28\%7=0$ , so the final answer is $\boxed{0}$

First off, crt dependents on relative Primes. 6 isn't Prime. 6 is equal to 2 * 3. Both prime. Rewrite problem as 2 congruent to x mod 3 because 2 sets of 2 is 4.

Now all mods are prime, so 7 5 3 = 105

Next substitute the mod in the method in the lesson. Get 7(10k + 5) or 70L + 35

35 is the number of interest, and 35 is a factor of 105, so no remainder. (105 = 3 35, 35 = 7 5)

Use the chinese remainder theorem

Let the number of coins be x.from the question we can have, X=6j +4 and j=5k +4 When substitute into the first equation, we get: X=30k+28 , if k=0 , then we have: X=28 and 7 divided by this number is 4 remainder 0 So,remainder equal 0

based on the given data, 6x+4=n; 5y+3=n; hence 6x-5y+1=0; to satisfy this equation put x=4 and y=5. so n=28. if its distributed among 7 reaming is 0.

According to the information, we get the two equations-5y+3=x and 6z+4=x where x is the no. of gold coins and z a y are two real numbers(I got the equations by divisor-remainder theorem).Solving the equations by trial and error method i got x=28.Now if we divide 28 by 7, naturally the remainder will be 0.

Let the number of coins = n. Then n = 6x+4 and n = 5y + 3. Where x & y are positive integer. So 6x + 4 = 5y + 3 or, 6x - 5y + 1 = 0, (x, y) = ( 5k - 1, 6k - 1) where k is integer. For k = 1, (x, y) = (4, 5). So smallest n is 28. 0 coin left when equally divided among seven friends?

So divided by 6 is 4 left: The possible amount of coins are: 4, 10, 16, 22, 28, 34, ... (4+6n) So divided by 5 is 3 left: The possible amount of coins are: 3, 8, 13, 18, 23, 28, 33, ... (3+5m) The smallest number that we get in both lists is 28. Now we divide 28 by 7: 28/7=4 with no rest. Therefor the solution is 0.

Total gold coins 28 If the coins are equally divided among six friends, four coins are left over it is 4 6=24+4=28 If the coins are equally divided among five friends, three coins are left over it is 5 5=25+3=28 If the coins are equally divided among seven friends, zero coins are left over it is 7*4=28+0=28 That's all

Now let the coins in the box was 28..... When 6 friends equally divided there 4 coins left...(28%6=4) When 5 friends divided there 3 coins left....(28%5=3) When 7 friends divided there 0 coins will left...(28%7=0)

1. (the smallest possible number of coins) divided by 6 = some number plus 4
2. (the smallest possible number of coins) divided by 5 = some number plus 3

Starting with requirement 2, any number that is divisible by 5 will end in a 5 or a 0, so any number that, when divided by 5, gives a remainder of 3, will end in a 3 or 8. Therefore we know that the number of coins will end in either a 3 or 8.

As for requirement 1, any number that is divisible by 6 will be even, and any number that when divided by 6 gives a remainder of 4 will be even. The number of coins must then be even.

Requirement 1 shows that the number of coins must end in a 3 or 8, and requirement 2 shows that the number of coins must be even, therefore the number of coins cannot end in a 3, it must end in an 8.

8 and 18 do not follow the requirements, but 28 does. 28 divided by 7 gives 4 with no remainder, so the solution to this problem is 0.

Let the smallest no. of gold coins in the box be x According to question, when x is divided by 6, 4 is the remainder. That means, ( x +2) is divisible by 6 According to question, when x is divided by 5, 3 is the remainder. That means, ( x +2) is divisible by 5 Hence, ( x +2) is the L.C.M. of 5 and 6 i.e. 30 So, x = 28 Now, when 28 is divided by 7, the remainder is 0. Therefore, no. of coins left when divided among 7 friends = 0

here no of coins can be written as, '6a+4 = 5b+3' = no of gold coins by putting different values of 'a', we can find the minimum value which is ''28''

I am using a very simple method Let no. of coins be x Acc. to the q., if we remove 4 and 6 from x respectively for the division by 6, the division will be complete. So:- so (x-4)/6 ϵ N also, (x-3)/5 ϵN

Using Hit and Trial, we get 28 to fulfill this condition being the smallest possible Natural Number [No. of coins can't be in decimals or negative].

So 28/7, gives 0 as the remainder :D .

In the problem, it says when the coins are equally divided among six people. four coins are left over, That means that the number of gold coins is four below or four above a multiple of six. Then, it says when the coins are equally divided among five people three coins are left over. That means that the number of coins is three below or three above a multiple of five. The smallest number meeting the two conditions is 28.

Let y is the total gold coins. So y = 6x + 4 and y = 5z + 3. We note z > x, x, y, z is positive intergers and 6x + 4 = 5z + 3 <=> x = (5z - 1)/6. We try with z = 1, 2, 3, 4, 5.... And the smallest value of z is 5 and x is 4 <=> y = 28 <=> 28/7 = 4 and the remainder is 0.