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If there are 50 notes whose total value is 100 rupees but 2 rupee note should not be there in the count of those 50 notes.How many such notes can be ?

Notes available are
$1$ Rupee
$2$ Rupees ( but NOT to use as per question )
$5$ Rupees
$10$ Rupees
$20$ Rupees
$50$ Rupees
$100$ Rupees ( obviously, this won't be needed!)
$500$ Rupees ( obviously, this won't be needed!)
$1000$ Rupees ( obviously, this won't be needed!)

Putting it other way

let us assume $a$,$b$,$c$,$d$ & $e$ as variables satisfying following condition :

1. $a+b+c+d+e = 50$
2. $a + 5b + 10c + 20d + 50e = 100$

Then what can be values of $a,b,c,d$ & $e$ ? ( Given that they must be non-negative integers!)

Thanks!

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5 Answers 5

up vote 7 down vote accepted

As you have written, we need to solve $$a+b+c+d+e = 50$$ and $$a+5b+10c+20d+50e = 100$$ where $a,b,c,d,e \in \mathbb{N}$. A good way in my opinion is to look at the problem on a case by case basis and solve.

It is better to start from $e$ and go all the way up to $a$ looking at all possible options for the values taken by $e,d,c,b,a$. Further note that $5 | a$.

A blind upper bound for the values taken by $a,b,c,d,e$ can be easily obtained.

$a \leq 50$, $b \leq 20$, $c \leq 10$, $d \leq 5$ and $e \leq 2$

Now we look at the different cases. There are not a lot of cases and in most of the cases we can easily reject it.

  • $e=2$ Not possible
  • $e=1$ We need $a+b+c+d = 49$ and $a+5b+10c+20d = 50$ Again not possible

Hence $e = 0$. Hence, $50$ rupee note is not needed.

  • $d=5$ Not possible
  • $d=4$ We need $a+b+c = 46$ and $a+5b+10c = 20$. Note that since $a,b,c \in \mathbb{N}$ we always need to have $a+5b+10c \geq a+b+c$. Hence, this case is also ruled out.
  • $d=3$ We need $a+b+c = 47$ and $a+5b+10c = 40$. Same argument as the previous case rules this out.
  • $d=2$ We need $a+b+c = 48$ and $a+5b+10c = 60$. Look at the possible sub cases now. Note that we need $a+5b \geq a+b$ and hence we need $60-10c \geq 48 -c$ and hence we need $c \leq \frac{4}{3}$ and hence $c \in \{0,1\}$
  • $d=2,c=0$ We need $a+b = 48$ and $a+5b = 60$. We get $a=45$ and $b=3$.
  • $d=2,c=1$ We need $a+b = 47$ and $a+5b = 50$. Not possible since $a,b \in \mathbb{N}$
  • $d=1$. In this case, we get $a+b+c = 49$ and $a + 5b+ 10c = 80$. Look at the possible subcases now. Note that we need $a+5b \geq a+b$ and hence we need $80 - 10c \geq 49 - c$ and hence $c \leq \frac{31}{9}$ and hence $c \in \{0,1,2,3\}$. Further $4b = (a+5b) - (a+b) = (80 - 10c) - (49-c) = 31 - 9c$. Hence $4 | (31-9c)$. Hence the only possibility is when $c = 3$
  • $d=1, c= 3$. We get $a+b = 46$ and $a+5b = 50$ which gives us $a = 45$ and $b = 1$
  • $d=0$. In this case, we need $a+b+c = 50$ and $a+5b+10c = 100$. Look at the possible subcases now. Note that we need $a+5b \geq a+b$ and hence we need $10 - 10c \geq 50 - c$ and hence $c \leq \frac{50}{9}$ and hence $c \in \{0,1,2,3,4,5\}$. Further $4b = (a+5b) - (a+b) = (100 - 10c) - (50-c) = 50 - 9c$. Hence $4 | (50-9c)$. Hence the only possibility is when $c = 2$
  • $d=0, c= 2$. We need $a+b =48$ and $a+5b = 80$. Solving this we get $a=40, b=8$

Hence, the only possible solutions are $$\{45,3,0,2,0\}$$ $$\{45,1,3,1,0\}$$ $$\{40,8,2,0,0\}$$

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1  
A bit messy and ad-hoc, but probably the simplest way to find the answer on your own. +1 –  Michael Burge Jun 2 '11 at 22:00

One way to think about this is that the average note has value 2 rupees and only one note is any lower. So each large note has to be paired with enough 1 notes to bring the average down to 2, and the group can be viewed as a "note". A 5 rupee note has to be paired with 3 1 rupee notes making a package that is worth 8 rupees and has 4 notes. Similarly a 10 rupee note must be paired with 8 1 rupee notes to make a package worth 18. Your new notes have value 8, 18, 38, 98 and you need to make 100 (the count will take care of itself). You can't use 98 at all. Working mod 8 can make it easy. If you use two 38's, you need 24 more, which can only be done with 3 8's. So this gives 2*20+3*5+45*1 (in the original notes). If you use one 38, that is 6 mod 8, so to get to 4 mod 8 we need 3 18's and 1 8, giving 20+3*10+5+45*1. Finally, if we use no 38's, we need 2 18's and 8 8's, which leads to 2*10+8*5+40*1.

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+1. Nice idea of combining notes! –  user17762 Jun 2 '11 at 20:34

Since you cannot use 2 Rupee notes, you must use lower values to counter the higher values. Since there is only one lower value (1 Rupee) and 4 higher values (5, 10, 20, 50), you need to balance how many 1 Rupee notes can be balanced out by which higher values to equal the same value for the same number of 2 Rupee notes. And how many of the higher values can be swapped to substitute how many of the smaller notes.

For example, replacing a 1 with a 5 saves you 4 notes, 5 with a 10 will allow you to save on 5 additional 1 Rupee notes, replacing a 5 with a 20 yields 15, etc.

If you then target a factor of 50 total notes, then it becomes a matter of simply repeating this value once you hit it.

The second highest factor of 50 is 25, which yields 50 Rupees in 2 Rupee notes. You'd need 50 1 Rupee notes to match that. You can replace 4 at a time with 5 Rupee notes to get 45 + 1*5 (46), 40 + 2*5 (42), 35 + 3*5 (38), 30 + 4*5 (34), 25 + 5*5 (30). At this point, you have 30 notes, which is only 5 away from the target of 25 notes. Since replacing a 5 Rupee note with a 10 Rupee note saves us exactly 5 notes, then we can get 25 notes totalling 50 Rupees with the combination of 20 + 4*5 + 1*10.

Since we started with the second highest factor, we only need to double the amount. 40 notes worth 1 Rupee, 8 notes worth 5 Rupees, and 2 notes worth 10 Rupees.

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$a = 5, b = 1, c = 2, d = 1, e = 1$ hence $a + 5b + 10c + 20d + 50e$ will be $5 + (5\times 1) + (10\times 2) + (20\times 1) + (50\times 1) = 5 + 5 + 20 + 20 + 50 = 100$.

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$a+b+c+d+e \neq 50$ –  Michael Albanese Oct 24 '13 at 5:46

1 rs note * 45 = 45

5 rs note * 3 = 15

20 rs note * 2 = 20

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