The document describes a "word math" problem where letters represent digits. Students are asked to find digit replacements for letters in word problems like FOUR + ONE = FIVE such that the sums are correct. There are constraints, like each letter must be a different digit and the first digit cannot be zero. The document provides methods and spreadsheets to systematically generate and check all possible solutions. It also addresses related common core math standards.

1. http://www.scctmprogram.org/2014-Sept-MathMate.pdf

2. “WORD MATH” RULES
Find which digits should replace each of the letters in each of
the following word sums. For each problem, each of the
letters stands for a different digit. The first digit in a number
is never zero. Replace the letters so that each sum is correct.
Can you find more than one solution? Are there restrictions on
which digits are possible for any of the letters? One of these
is impossible. Explain how you know.

3. “WORD MATH” RULES
Find which digits should replace each of the letters in each of
the following word sums. For each problem, each of the
letters stands for a different digit. The first digit in a number
is never zero. Replace the letters so that each sum is correct.
Can you find more than one solution? Are there restrictions on
which digits are possible for any of the letters? Explain how
you know.
+

4. REDUCING THE PROBLEM
F O U R
+ O N E
F I V E
• R + E = E, so R = 0 (identity property of
addition)
• O + O = I with no carry, so O = 1, 2, 3, 4

5. ASSISTING STUDENTS’ WORK ON PROBLEM

6. SCAFFOLDING SPREADSHEET #1

7. SCAFFOLDING SPREADSHEET #2

8. CCSS CONTENT STANDARDS ADDRESSED
Students should :
• Understand the difference between face value
and place values of digits in multi-digit numbers
(2.NBT.1, 5.NBT.1)
• Use their place value understanding and
properties of operations to perform multi-digit
arithmetic (3.NBT.2, 4.NBT.4)
• Evaluate expressions in which letters stand for
numbers (6.EE.2)
• Determine whether a given number makes an
equation true (6.EE.5)
• Understand that a variable can represent an
unknown number or any number in a specified
set (6.EE.6).

9. CCSS STANDARDS FOR MATHEMATICAL PRACTICE ADDRESSED
Make sense of problems and persevere in
solving them
 Understand what the problem is asking –
what is the nature of the solutions which
exist?
Construct viable arguments & critique the
reasoning of others
 Describe the constraints on variables.
 Listen to others’ solution methods
 Evaluate efficiencies of different solution
methods.
Attend to precision
 Make sure the numbers add correctly.
 Make sure that each different letter is
represented by a different digit.
 Keep track of the digits they have already
used & the digits that are remaining.
 Find ways to record their valid & invalid
solutions.
Look for and express regularity in repeated
reasoning
 Look for patterns.
Reason abstractly and quantitatively
 Substitute letters for digits.
 Understand that the digit chosen for each
letter is a face value.
 Understand the place value of the digit &
its corresponding letter.
Use appropriate tools strategically
 Decide what tools to use
Look for & make use of structure
 Create a system to record addends which
result in a carry.
 Determine when a carry to the next higher
place value will be used.
Model with mathematics
 Letter-number substitution problems are
used in cryptography, a real-world
application.

10. EXTENSION QUESTIONS
Question 1: Did you notice any shortcuts to finding
solutions, and if so, why are these valid shortcuts?
Question 2: Given a solution to the FOUR+ONE=FIVE
problem, 5130 + 167 = 5297, find as many different
solutions as you can using the same digits. How will you
know when you have found all of the possible solutions
using these digits?
Question 3: Find the total number of solutions there are to
this word sum. Describe how you found this number of
solutions.
Question 4: Answer the question, “If I find all of the
solutions to this problem for F = 1, can I multiply that
number of solutions by 9 to find the total number of
solutions to this problem? Why or why not?”
Question 5: Create your own letter-number substitution

11. SOLUTION METHOD #1: PLACE VALUES
F1O1UR = F*1000+O1*100+U*10+R O2NE = O*100+N*10+E
F O1 U R
1000 100 10 0
2000 200 20
3000 300 30
4000 400 40
5000 50
6000 60
7000 70
8000 80
9000 90
O2 N E
100 10 1
200 20 2
300 30 3
400 40 4
50 5
60 6
70 7
80 8
90 9

12. DISTINCT DIGITS
FOR
F1, O1, U, R
(219 SETS)
F1 = 1 F1 = 2 F1= 3 F1 = 4 F1 = 5 F1 = 6 F1 = 7 F1 = 8 F1 = 9
1230 2130 3120 4120 5120 6120 7120 8120 9120
1240 2140 3140 4130 5130 6130 7130 8130 9130
1250 2150 3150 4150 5140 6140 7140 8140 9140
1260 2160 3160 4160 5160 6150 7150 8150 9150
1270 2170 3170 4170 5170 6170 7160 8160 9160
1280 2180 3180 4180 5180 6180 7180 8170 9170
1290 2190 3190 4190 5190 6190 7190 8190 9180
1320 2310 3210 4210 5210 6210 7210 8210 9210
1340 2340 3240 4230 5230 6230 7230 8230 9230
1350 2350 3250 4250 5240 6240 7240 8240 9240
1360 2360 3260 4260 5260 6250 7250 8250 9250
1370 2370 3270 4270 5270 6270 7260 8260 9260
1380 2380 3280 4280 5280 6280 7280 8270 9270
1390 2390 3290 4290 5290 6290 7290 8290 9280
1420 2410 3410 4310 5310 6310 7310 8310 9310
1430 2430 3420 4320 5320 6320 7320 8320 9320
1450 2450 3450 5340 6340 7340 8340 9340
1460 2460 3460 5360 6350 7350 8350 9350
1470 2470 3470 5370 6370 7360 8360 9360
1480 2480 3480 5380 6380 7380 8370 9370
1490 2490 3490 5390 6390 7390 8390 9380
5410 6410 7410 8410 9410
5420 6420 7420 8420 9420
5430 6430 7430 8430 9430
5460 6450 7450 8450 9450
5470 6470 7460 8460 9460
5480 6480 7480 8470 9470
5490 6490 7490 8490 9480

13. O2 = 1 O2= 2 O2= 3 O2= 4
120 165 201 256 301 356 401 456
123 167 203 257 302 357 402 457
124 168 204 258 304 358 403 458
125 169 205 259 305 359 405 459
126 170 206 260 306 360 406 460
127 172 207 261 307 361 407 461
128 173 208 263 308 362 408 462
129 174 209 264 309 364 409 463
130 175 210 265 310 365 410 465
132 176 213 267 312 367 412 467
134 178 214 268 314 368 413 468
135 179 215 269 315 369 415 469
136 180 216 270 316 370 416 470
137 182 217 271 317 371 417 471
138 183 218 273 318 372 418 472
139 184 219 274 319 374 419 473
140 185 230 275 320 375 420 475
142 186 231 276 321 376 421 476
143 187 234 278 324 378 423 478
145 189 235 279 325 379 425 479
146 190 236 280 326 380 426 480
147 192 237 281 327 381 427 481
148 193 238 283 328 382 428 482
149 194 239 284 329 384 429 483
150 195 240 285 340 385 430 485
152 196 241 286 341 386 431 486
153 197 243 287 342 387 432 487
154 198 245 289 345 389 435 489
156 246 290 346 390 436 490
157 247 291 347 391 437 491
158 248 293 348 392 438 492
159 249 294 349 394 439 493
160 250 295 350 395 450 495
162 251 296 351 396 451 496
163 253 297 352 397 452 497
164 254 298 354 398 453 498
Distinct
digits for
O2, N, E
(280 sets)

14. SUMS FOR
EACH
COMBINATION
OF F1O1UR
AND O2NE
(219 * 280 =
61,320 SETS)
O2NE
120 123 124 125 126
F1O1UR
1230 1230+120=1350 1230+123=1353 1230+124=1354 1230+125=1355 1230+126=1356
1240 1240+120=1360 1240+123=1363 1240+124=1364 1240+125=1365 1240+126=1366
1250 1250+120=1370 1250+123=1373 1250+124=1374 1250+125=1375 1250+126=1376
1260 1260+120=1380 1260+123=1383 1260+124=1384 1260+125=1385 1260+126=1386
1270 1270+120=1390 1270+123=1393 1270+124=1394 1270+125=1395 1270+126=1396
1280 1280+120=1400 1280+123=1403 1280+124=1404 1280+125=1405 1280+126=1406
1290 1290+120=1410 1290+123=1413 1290+124=1414 1290+125=1415 1290+126=1416
1320 1320+120=1440 1320+123=1443 1320+124=1444 1320+125=1445 1320+126=1446
1340 1340+120=1460 1340+123=1463 1340+124=1464 1340+125=1465 1340+126=1466
1350 1350+120=1470 1350+123=1473 1350+124=1474 1350+125=1475 1350+126=1476
1360 1360+120=1480 1360+123=1483 1360+124=1484 1360+125=1485 1360+126=1486
1370 1370+120=1490 1370+123=1493 1370+124=1494 1370+125=1495 1370+126=1496
1380 1380+120=1500 1380+123=1503 1380+124=1504 1380+125=1505 1380+126=1506
1390 1390+120=1510 1390+123=1513 1390+124=1514 1390+125=1515 1390+126=1516

15. LOGICAL STATEMENT CHECKS
For each of 61,320 possibilities for “F1O1UR +
O2NE = F2IVE,” check 10 logic statements:
 Whether F1 is the same as F2
 Whether O1 is the same as O2
 Whether F1 is distinct from N, E, I, V
 Whether O1 is distinct from N, E, I, V
 Whether U is distinct from N, E, I, V
 Whether R is distinct from N, E, I, V
 Whether N is distinct from I, V
 Whether E is distinct from I, V
 Whether I is distinct from V
If and only if all of the above are true, then the solution is valid
for “FOUR + ONE = FIVE.”

16. LOGICAL STATEMENT CHECKS – SCREEN SHOT

17. CONCLUDING SPREADSHEET SNAPSHOT

18. Observation #1
F and E do not depend on the other letters. There are 36 pairs of 2-digit
combinations:
 1,2 2,3 3,4 4,5 5,6 6,7 7,8 8,9
 1,3 2,4 3,5 4,6 5,7 6,8 7,9
 1,4 2,5 3,6 4,7 5,8 6,9
 1,5 2,6 3,7 4,8 5,9
 1,6 2,7 3,8 4,9
 1,7 2,8 3,9
 1,8 2,9
 1,9
The possibilities for F and E are interchangeable.
METHOD 2: COMBINATORICS #1 F O U R
+ O N E
F I V E

19. Observation #2
O + O = I with no carry. Thus, O can only be 1, 2, 3, or 4.
Observation #3
U + N = V can, but does not need to, have a carry.
Observation #4
The possibilities for U and N are interchangeable (commutative property of
addition).
* For every combination of F, E, U, N, we can wrote 4 possible
arrangements: FEUN, EFUN, FENU, EFNU. Thus the total # of possible
solutions is a multiple of four.
F O U R
+ O N E
F I V E

20. METHOD FOR GENERATING SOLUTIONS – DISTINCT DIGITS
For each of the 36 pairs of (F,E) possibilities, find O, I, U, N, V possibilities:
O + O = I No Carry from U + N = V
1+1=2 1+2=3 2+3=5 3+4=7 4+5=9
2+2=4 1+3=4 2+4=6 3+5=8
3+3=6 1+4=5 2+5=7 3+6=9
4+4=8 1+5=6 2+6=8
1+6=7 2+7=9
1+7=8
1+8=9
O + O = I Carry from U + N = V
1+1  3 2+9 1 4+8  2 5+8 3 6+9 5
2+2  5 3+8 1 4+9  3 5+9 4 7+8 5
3+3 7 3+9 2 5+6  1 6+7 3 7+9 6
4+4 9 4+7 1 5+7  2 6+8 4 8+9 7
There are 300
solutions for (F, E,
O, I, U, N, V) with
F<E and U<V, so
there are 300* 4 =
1200 total
solutions

21. METHOD 3: COMBINATORICS #2
• For every combination of (O, I, V), there are 2 combinations
of (U,N).
• For every combination of (O, I, V, U, N, R=0), there are 12
combinations of (F, E):
* Since O, I, V, U, N, R represent 6 distinct digits, there are 4
remaining digits for F and E
* For each of the 4 digits that are possible for the value of
F, there are 3 remaining digits possible for the value of E.
(4 * 3 = 12)
 If we find the number of combinations of
(O, I, U, N, V, R) where U < N,
we have 2*12 = 24 times more
total solutions to (F, E, O, I, U, N, V, R)
F O U R
+ O N E
F I V E

22. 50 COMBINATIONS OF
(O, U, N, I, V)
O U N I V O U N I V O U N I V O U N I V
1 3 4 2 7 2 1 5 4 6 3 1 4 6 5 4 1 2 8 3
1 3 5 2 8 2 1 6 4 7 3 1 7 6 8 4 1 5 8 6
1 3 6 2 9 2 1 7 4 8 3 1 8 6 9 4 1 6 8 7
1 4 5 2 9 2 1 8 4 9 3 2 5 6 7 4 2 3 8 5
1 4 8 3 2 2 3 5 4 8 3 2 7 6 9 4 2 5 8 9
1 5 7 3 2 2 3 6 4 9 3 2 9 7 1 4 2 7 8 9
1 5 9 3 4 2 3 8 5 1 3 4 5 6 9 4 3 6 8 9
1 6 8 3 4 2 4 7 5 1 3 4 8 7 2 4 3 8 9 1
1 6 9 3 5 2 4 9 5 3 3 5 6 7 1 4 5 6 9 1
1 7 8 3 5 2 6 7 4 3 3 5 9 7 4 4 5 7 9 2
1 7 9 3 6 2 6 8 5 4 3 6 8 7 4 4 5 8 9 3
1 8 9 3 7 2 7 9 5 6 3 6 9 7 5 4 6 7 9 3
2 8 9 5 7 4 7 8 9 5
F O U R
+ O N E
F I V E
50 * 24 = 1200 TOTAL SOLUTIONS