I am a number less than 3000. When you divide me by 32, my remainder is 30. When you divide me by 58, my remainder is44. What number am I?

Question

My son is doing 5th grade math & I don't know how to solve it

 

Answer 1

First, list all of the multiples of 32 (that is, 32*1, 32*2, 32*3, …) that are less than 3000.

List #1:   32 64 96 128 160 192 224 256 288 320 352 384 416 448 480 512 544 576 608 640 672 704 736 768 800 832 864 896 928 960 992 1024 1056 1088 1120 1152 1184 1216 1248 1280 1312 1344 1376 1408 1440 1472 1504 1536 1568 1600 1632 1664 1696 1728 1760 1792 1824 1856 1888 1920 1952 1984 2016 2048 2080 2112 2144 2176 2208 2240 2272 2304 2336 2368 2400 2432 2464 2496 2528 2560 2592 2624 2656 2688 2720 2752 2784 2816 2848 2880 2912 2944 2976

 

Then, add 30 (the remainder after division) to each of these:

 

List #2:   62 94 126 158 190 222 254 286 318 350 382 414 446 478 510 542 574 606 638 670 702 734 766 798 830 862 894 926 958 990 1022 1054 1086 1118 1150 1182 1214 1246 1278 1310 1342 1374 1406 1438 1470 1502 1534 1566 1598 1630 1662 1694 1726 1758 1790 1822 1854 1886 1918 1950 1982 2014 2046 2078 2110 2142 2174 2206 2238 2270 2302 2334 2366 2398 2430 2462 2494 2526 2558 2590 2622 2654 2686 2718 2750 2782 2814 2846 2878 2910 2942 2974 3006 (note: eliminate 3006 because it is not less than 3000)

Now, list all the multiples of 58 up to 3000:
List #3:   58 116 174 232 290 348 406 464 522 580 638 696 754 812 870 928 986 1044 1102 1160 1218 1276 1334 1392 1450 1508 1566 1624 1682 1740 1798 1856 1914 1972 2030 2088 2146 2204 2262 2320 2378 2436 2494 2552 2610 2668 2726 2784 2842 2900 2958

And, add 44 to each of these:

List #4: 102 160 218 276 334 392 450 508 566 624 682 740 798 856 914 972 1030 1088 1146 1204 1262 1320 1378 1436 1494 1552 1610 1668 1726 1784 1842 1900 1958 2016 2074 2132 2190 2248 2306 2364 2422 2480 2538 2596 2654 2712 2770 2828 2886 2944 3002 (note: eliminate 3002)

The number(s) that satisfy the problem requirements are in both List #2 and List #4. [The easy way to search is to take the shorter list (List #4) and determine whether each number is in the other list (List #2).]

Here is a computer solution to the same problem:

FOR I GOES FROM 1 to 3000
    IF ( ((I MOD 32)=30) AND ((I MOD 58)=44) ) THEN OUTPUT I
NEXT I

The numbers are: 798, 1726, and 2654

Answer 2

Suppose you are X. Then:

X = 32 Y + 30 
X = 58 Z + 44

where Y and Z are the integer numbers. Notice, Y ≤ 93, and Z ≤ 51 (this comes due to the fact that X cannot be larger than 3000). 

Now setting the above equation equal you would find:

32 Y + 30 = 58 Z + 44
or
32 Y = 58 Z + 14
16 Y = 29 Z + 7

Subtract 16 Z from both sides:

16(Y-Z) = 13 Z + 7

Since both Y and Z are integers, Y-Z is also an integer, let us call it W, where W < 42 (coming from 93 and 51).

16 W = 13 Z + 7 
or
16 W - 7 = 13 Z

Z = (16 W - 7) / 13 = W + (3 W - 7) / 13

Subtract both sides by W:

Z - W = (3 W - 7) / 13

Call Z - W with A, again A is an integer. A < 9. 

13 A = 3 W - 7
3 W = 13 A + 7
W = (13 A + 7)/3 = 4 A + (A+7)/3. 

Subtract both sides by 4A.

W - 4A = (A+7)/3

Again, W - 4A is an integer, call this B. B < 6. 

B = (A+7)/3
3 B = A + 7
A = 3B - 7

Now the problem has become so simple: For all values of B from 0 to 6 find a solution that 3B - 7 is a positive integer. Checking is easy:

B = 1, A = 3B - 7 = -4 ---> Does not work!
B = 2, A = 3B - 7 = -1 ---> Does not work!
B = 3, A = 3B - 7 = 2 ---> Works, W = (13 A + 7)/3 = 11
B = 4, A = 3B - 7 = 5 ---> Works, W = (13 A + 7)/3 = 24
B = 5, A = 3B - 7 = 8 ---> Works, W = (13 A + 7)/3 = 37

Notice there are three solutions. We will try to reconstruct now for Y and Z. 

Z = (16 W - 7) / 13 = (16*11 - 7) / 13 = 13
Z = (16 W - 7) / 13 = (16*24 - 7) / 13 = 29
Z = (16 W - 7) / 13 = (16*37 - 7) / 13 = 45

Corresponding Y values are 

Y = (29 Z + 7)/16 = (29*13 + 7)/16 = 24
Y = (29 Z + 7)/16 = (29*29 + 7)/16 = 53
Y = (29 Z + 7)/16 = (29*45 + 7)/16 = 82

Thus we have found three pairs of solutions:

(Y, Z) = (24, 13)
(Y, Z) = (53, 29)
(Y, Z) = (82, 45)

Indeed, if you check they all satisfy the first equation as well as the condition. 

Correspondingly, the missing number would be:

• X = 798
• X = 1726
• X = 2654

Thus we have found three solutions! Some people go through trial and errors, which is not the way a mathematician should do. 

Answer 3

Seeing that this is 5th grade math, take the following approach:

when there is a remainder of 30 your answer divided by 32 will be AB.9375 where AB is the whole number part

when there is a remainder of 44 your answer will be AB.7586207...

3000/58=51.7

51*58=2958

2958+44=3002  too large

50*58=2900

2900+44=2944

2944/58=50.7586207...

2944/32=92 which is not what we want (we want AB.9375)

now decrease 2944 by multiples of 58 until you get the number you want; the remainder of 44 is already added to a multiple of 58, namely 2900 so you don't have to be concerned about 44 anymore

2944-58=2886

2886-58=2828

2828-58=2770

2770-58=2712

2712-58=2654  this is one of the numbers; the others won't work because I tried them

2654/32=82.9375 which is what you want

2654/58=45.7586207... which is what you want

continue the process to see if there are any other numbers

Farrooh's solution shows that there are 2 more numbers that satisfy the conditions