The Ingenuity Of An Army Clerk

As the total number of items is an even number, and all the items come in piles of even numbers except for rifles, you know that there is an even number of piles of rifles (minimum number 210).

Also, as the total number is 1126, you know that the number of uniforms must end in 6 (Minimum of 336 uniforms), since all the other items (rifles included) come in multiples of ten.

So the minimum number of rifles is 210, minimum number of uniforms 336, minimum number of berets 140, and minimum number of pairs of boots 60. (Total 746)

As there aren't enough rifles in this situation, consider that there are 420 instead. This would mean a total of 956 iteams, with 170 still to go. However as no comination of 84, 140 and 60 adds up to 170, there cannot be 420 rifles. Instead consider there are 630 rifles, this would mean the total is 1166 items, which is too many. Therefore there cannot be more than 210 rifles, which is not enough.

Let $x,y,z,w$ be the number of units of berets, rifles, uniforms, and boots, respectively. We get $140 x + 105 y + 84 z + 60 w = 1126$ . Looking mod $3,5,7$ , we get $x \equiv 2$ mod $3$ , $z \equiv 4$ mod $5$ , $w \equiv 5$ mod $7$ . Writing $x = 2 + 3x_0$ , $z = 4 + 5 z_0$ , $w = 5 + 7 w_0$ , we get $420 x_0 + 105 y + 420 z_0 + 420 w_0 = 210.$ Since all the variables are nonnegative integers, it's clear by inspection that the only solution is $y = 2$ and all the others are $0$ .

So there is a unique solution $(x,y,z,w) = (2,2,4,5)$ in nonnegative integers. There are 280 berets, 210 rifles, 336 uniforms, and 300 pairs of boots. That's not enough rifles.

Okay guys this is how I cracked it

Since we are given a total number of 1126 individual items, first we find the total number of berets , rifles , uniforms and boots that make up this total: that is we find the number of stacks that each contain 140 berets , the number of piles that each contain 105 rifles , the number of crates that each contain 84 uniforms and the number of boxes that each contain 60 pairs of boots .

Is almost a tedious process to figure out the exact number of stacks, piles, crates and boxes of berets, rifles, uniforms and boots respectively but on closer examination, we can do this;

The only single digit numbers that gives a unit digit of 6 when they multiply 4 are 9 and 4. Thus the possible number of crates of uniforms are 4 or 9. On doing trial and error, 9 is too big so the only possible number of crates of uniforms each containing 84 uniforms are 4.

2 in this case is the smallest possible number that can multiply 5 and gives a unit digit of 0. Thus the possible number of piles of rifles each containing 105 guns are 2.

This leaves us with the berets and boots.

Adding the total number of rifles and uniforms gives 546 and taking this sum from the total number of individual items we are left with 580 individual items.

Clearly, to get a sum of 580 between the berets and the boots, we simply multiply the number of berets in each stack by 2 and the number of pairs of boots in each box by 5. Thus we have;

The number of berets in each stack equals ${2}\times{140}={280}$

The number of rifles in each pile equals ${2}\times{105}={210}$

The number of uniforms in each crate equals ${4}\times{84}={336}$

The number of pairs of boots in each box equals ${5}\times{60}={300}$

These sums add up to 1126 individual items .

Clearly there are less rifles to outfit all the 250 recruits .

Hence the correct answer is * No, they do not have enough rifles * .

First let the number of berets be = a , number of rifles be = b, number of uniform be = c and number of boots =d.

Now as per given in the question, we get an eqn : 140a + 105b + 84c + 60d = 1126

here, the mod represents : a= 2mod3 d= 5mod7 c= 4mod5 ( put a= 2+3x' ; c=4+5c' ; d=5+7d')

Final Eqn: 420a' + 105b + 420c' + 420d' = 210

If we put b=2, all others will become zero.

[a,b,c,d] = [2,2,4,5] -> Results : 336 uniforms, 300 boots, 210 rifles & 280 berets.
-> Therefore, there are not many rifles !

they have 336 uniforms(84 X 4) , 300 boots(60 X 5) , 280 berets(140 X 2) , and 210 refiles (105 X 2) , which totals 1126 individual items. Therefore they have 40 refiles less.

I did this by writing a table of multiples of 140, 105, 84 and 60 (corresponding to berets, refiles, uniforms and boots respectively). Since the total of items is exactly 1126, multiples of these four numbers must add up to 1126. We immediately know that there cannot be an odd number piles of rifles, since 105 is odd (the unit digit would be odd and never 6 of 1126). Therefore there must be either 2 or 4 piles of rifles. If there were 4 piles, there would be $105\times4 = 420$ , which doesn't leave much for the other items, and there is no possible combination to make 1126. So there must be 2 piles of rifles. Since the entire company has 250 recruits, and $105\times2 = 210$ ; the answer is "No, they do not have enough rifles."

If all items are sufficient for the whole company, there would be:

• (250/140) = 1.785 ~ 2 stacks of berets
• (250/105) = 2.381 ~ 3 piles of rifles
• (250/84) = 2.976 ~ 3 crates of uniforms
• (250/60) = 4.167 ~ 5 boxes of boots

In that case,

Number of items = (140x2) + (105x3) + (84x3) + (60x5) = 1147 items

This exceeds the total number of items available by (1147-1126) = 21 units.

Looking through the figures, we find that if we increase the number of uniform crates by 1 and reduce the number of rifle piles by 1, the total number of items will be reduced by 21 units.

Checking:

Number of items = (140x2) + (105x2) + (84x4) + (60x5) = 1126 items

Therefore the answer is "they don't have enough uniform".

Don't get bewildered by a wordy problem; apply simple Maths.

As each of 250 recruits is to be provided with a beret, a rifle, a uniform, and a pair of boots, minimum No. of Individual Items (denoted by II hereafter) needed is 1000.

Now, notice the last digit of 1126 ( $\boxed{6}$ ). We can deduce that No. of Uniform-Crates is 4 so that we land up having $\boxed{6}$ at Unit's place.

We could have chosen 9 Uniform-Crates even, but the problem is the II . 9 Uniform-Crates $\rightarrow$ 756 uniforms $\Rightarrow$ ( 1126 - 756 = 370 left II ). And Certainly, we can't see the recruits without rifles or even boots! Can we? (Minimum II needed is 750, condition met by 4 Uniform-Crates ONLY)

Now, we are left with (1126 - (4 $\times$ 84)) = 790 II

With 790 II we can't distribute the supplies to 280 (needing 840) and 300 (needing 900) recruits. So, Option 2 and Option 4 can be eliminated. Option 1 stands incorrectly as we've already successfully distributed the uniforms. So, Option 3 is the correct answer because no. of rifles is insufficient.

we should have 2 stacks of berets, 2 piles of rifles, 4 crates of uniforms and 5 boxes of shoes. to get the total number of items as 1126. thus there are 280 berets, 210 rifles, 336 uniforms and 300 pairs of boots and hence they donot have enough rifles.

We have sets of berets (call them BER: 1 BER = 140), rifles (RIF), uniforms (UNI) and boots (BOT). The total number is 1126 i.e. w * BER + x * RIF + y * UNI + z * BOT = 1126. Here, BER = 140, RIF = 105, UNI = 84, BOT = 60.

We need to start eliminating the variables w, x, y, z etc. We'll use modulo arithmetic for this. As you can see, RIF, UNI and BOT are divisible by 3 but BER is not. If you divide BER by 3 you get 2 as a remainder.

w * BER mod 3 + x * RIF mod 3 + ... = 1126 mod 3 (divide both sides by 3 and calculate the remainders)

=> (w * 2) mod 3 = 1 (RIF, UNI and BOT all have 0 after we take modulo 3)

=> w = 2, 5, 8... 8 is impossible since then all the items except 6 will become berets! (And the minimum stack size is 60, for boots.) Only possibilities are then w = 2, 5.

First, we can check if we get a fit solution for x, y, z with w = 2. We can eliminate x by mod 6 and then y by mod 10.

One possible solution then becomes: 2 BER, 2 RIF, 4 UNI, 5 BOT. There may be other solutions but we've already found one possible case where there are not enough items; in this case there are only 2 * 105 = 210 rifles.

So we know which answer to pick!

no, is confirm so you start thinking for only two option....you will easily get answer

To solve this problem,u should first see the total number of individual items available. i.e. 1126 since the unit's place has 6 in individual items and 0,5,4 and 0 in case of stacks of beret,rifle,uniform and boots Now let us try to reduce the individual number of items,for that 1126 has 6 in units place so considering uniforms it has to be a multiple of 4 or 7 stacks.4 is the best choice,which leaves 1126-376 items. Now we have 750 items.Lets try to reduce this number again..So we take an account of boots for minimum of 250,we must multiply number of boot stacks by 5,which leaves 750-300=450 individual items So the equation now becomes 140x+105y=450. To satisfy 280 of them,x must be multiple of 2,which leaves insufficient number of rifles .