Buying ingings

Suppose Calvin buys $a$ bingbings, $b$ dingdings, $c$ fingfings and $d$ hinghings, then $$30a + 42b + 70c + 105d = 593 \tag{1} $$ is a linear diophantine equation, which describes the situation, where $a,b,c,d \in \mathbb{N} \cup \{0\}$ . This equation may look scary, but it is very nice actually, it is important to note that there are many common factors at some terms, for example $7 \mid 42$ , $7 \mid 70$ , $7 \mid 105$ , hence considering the equation $(1)$ modulo $7$ we get $$2a \equiv 5 \pmod{7} \Leftrightarrow a \equiv 6 \pmod{7} \tag{2}$$ Similarly we get $$2b \equiv 3 \pmod{5} \Leftrightarrow b \equiv 4 \pmod{5} \tag{3}$$ $$c \equiv 2 \pmod{3} \tag{4}$$ These three congruences are equivalent to saying that there exist integers $x,y,z$ such that $$a = 7x + 6 $$ $$b = 5y + 4$$ $$c = 3z + 2$$ So from equation $(1)$ we get $$d = 1 - 2x - 2y - 2z$$ Thus the equation $(1)$ has solution $$(a,b,c,d) = (7x + 6,5y + 4,3z + 2, 1 - 2x - 2y - 2z)$$ Remember that we need $a,b,c,d \in \mathbb{N} \cup \{0\}$ , if $x < 0$ , then $a < 0$ , which we can't have, if $y < 0$ , then $b < 0$ and if $z < 0$ , then $c < 0$ , hence we need $$a,b,c \geq 0$$ however if one of $a,b,c$ is greater than $0$ , then $d < 0$ , which we can't have, hence $$x = y = z = 0$$ and we easily get $$(a,b,c,d) = (6,4,2,1)$$ as only possible solution, hence $$a + b + c + d = 6 + 4 + 2 + 1 = \boxed{13}$$

We first notice the tenth digit in 59.30. The tenth digit is 3 which is odd. The only number contributing to the decimal points are 10.50(hinghings) and 4.20(dingdings). Therefore the sum of a combinations of them produces such a number which has 3 in its tenth digit. The only ones are: 1. 10.50(3) + 4(4.20)= 48.3
2. 10.50(1) + 9(4.20)= 48.3
3. 10.50(1) + 4(4.20)= 27.3
The first 2 ones produces 48.30 which when subtracted from 59.30 is equal to 11. Therefore the sum of a combination of 7 and 3 must produce 11. Namely, 3x+7y =11; this equation we clearly see has no integer solution. Therefore we consider the last case when the sum is equal to 27.3. If we subtract it from 59.30 it gives 32.59. 59.30 - 27.30= 32.00 Now the equation is 3x +7y= 32. We can easily see that equation holds true when x=6 and y=2. Therfore the number of bingbings is 6, digdings 4, fingfings 2 and hinghings 1. As $3(6) + $4.20(4) + $7(2) + $10.50(1) is indeed equal to $59.30. So the answer is 6+4+2+1=13

Here first one needs to figure out that how (d) & (b) adjust the first decimal place of the total sum 59.30.$ The very first possibility is taking 4 (d) & 1(h) which sets the first decimal place 0.3.and a total sum 27.30$. Rest of the job is very easy: just to adjust 32$ with (b) & (f). So 6(b) & 2(f) yields rest of the amount; i.e. 32$ So the final answer is.....13 items

a to make total 59.3 we have to eliminate the decimal factor....so minimum posssible combination to remove decimal fector is

(59.3)-(10.5)-4(4.2)=32 [if 32 is not satisfied wid any combination of other two items then we try the another combination of of first two items to remove decimal side] now 32=2(7)-6(3) [ we are lucky we got the combination in first chance so total items 1+4+2+6=13

6 X 3 = 18,00; 4 X 4,20 = 16,80; 2 X 7,00 = 14,00; 1 X 10,50 = 10,5; Soma dos itens 6 + 4 + 2 + 1 = 13.

Soma do dinheiro: 18,00 + 16,80 + 14,00 + 10,50 = 59,30

Let $b,d,f,h$ be the number of bingbings, dingdings, fingfings, and hinghings that Calvin buys. Then $30b + 42d + 70f + 105h = 593$ It must be the case that $d = 5d' + 4$ and $h = 2h' + 1$ for some non-negative $d', h'$ because the last digit of the LHS must equal the last digit of the RHS (i.e. 3). Substituting for $d,h$ gives $30b + 70f + 210 (d' + h') = 320$ When $d' + h' = 1, 30b + 70f = 110$ , which has no solution. It must be that $d' + h' = 0$ and $f = 2, b = 6$ satisfies the equation. Therefore, $b = 6, d = 4, f = 2, h = 1$ and their sum is $\fbox{13}$

Calvin bought 6 bingbings(6 3=18);4 dingdings (4.20 4=16.8) ;2 fingfings(2 7=14) 1 hinghings (1 10.5=10.5) ,so you get amount total=18+16.8+14+10.5=59.30; items total= 6+4+2+1=13. the key to solve this problem is total decimal is .30.so first find for it, and then fix other items.

in the resultant .30 has to be notices . it can only by 10.50 and using 4.20 its comes 1 $10.50 and 4 $4.20 then leftover is $32 it can be brought by 2 items $3 ans $7 and using a little logic we find that its 6 $3 and 2 $7

In order to arrive at a 30 cent total we need (2n+1) hinghings (to arrive at 50 cents) and (5m+4) dingdings (to arrive at 30 cents).

{n=0, m=0} gives $27.30

{n=0, m=1} gives $48.30

{n=1, m=0} gives $48.30

{n=1, m=1} gives $69.30, which has exceeded our total of $59.30.

We can intuitively reject {n=0, m=1} and {n=1, m=0}. If either of them resulted in a valid solution, the other must certainly result in a valid solution, but the total number of items he bought would be different. Hence we know he bought 4 dingdings and 1 hinghing.

We need to split up the remainder of $32 between bingbings ($3) and fingfings ($7).

We fill up our inventory with fingfings without going over $32, and attempt to fill up the remainder with bingbings.

4 fingfings = $28, requires 1.33 bingbings.

3 fingfings = $21, requires 3.67 bingbings.

2 fingfings = $14, requires 6 bingbings.

Hence solution is 6 bingbings, 4 dingdings, 2 fingfings and 1 hinghing.

Since the total is 59.3, we need to make use of the items with prices having decimals.

(59.3)-(10.5)-4(4.2)=32

(we subtract 5 items to make the total an integer for us to use the remaining integer prices)

So,

32-2(7)-6(3)=0

(here, we use 8 items)

In total, Calvin buy 5+8=13 items.