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

CRYPTARITHMETIC 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.

Are these possible? If so, can you find more than one solution? Are there restrictions on which digits are possible for any of the letters?

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).

CCSS STANDARDS FOR MATHEMATICAL PRACTICE ADDRESSEDMake 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 useLook 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.

W I N + T H ER A C E

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.

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.

+

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

ASSISTING STUDENTS’ WORK ON PROBLEM

SCAFFOLDING SPREADSHEET #1

Blackboard > Content > FOUR ONE FIVE > Spreadsheet

SCAFFOLDING SPREADSHEET #2

EXTENSION QUESTIONSQuestion 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 problem and find its solution(s).

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 E100 10 1200 20 2300 30 3400 40 4

  50 5  60 6  70 7  80 8  90 9

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 91201240 2140 3140 4130 5130 6130 7130 8130 91301250 2150 3150 4150 5140 6140 7140 8140 91401260 2160 3160 4160 5160 6150 7150 8150 91501270 2170 3170 4170 5170 6170 7160 8160 91601280 2180 3180 4180 5180 6180 7180 8170 91701290 2190 3190 4190 5190 6190 7190 8190 91801320 2310 3210 4210 5210 6210 7210 8210 92101340 2340 3240 4230 5230 6230 7230 8230 92301350 2350 3250 4250 5240 6240 7240 8240 92401360 2360 3260 4260 5260 6250 7250 8250 92501370 2370 3270 4270 5270 6270 7260 8260 92601380 2380 3280 4280 5280 6280 7280 8270 92701390 2390 3290 4290 5290 6290 7290 8290 92801420 2410 3410 4310 5310 6310 7310 8310 93101430 2430 3420 4320 5320 6320 7320 8320 93201450 2450 3450   5340 6340 7340 8340 93401460 2460 3460   5360 6350 7350 8350 93501470 2470 3470   5370 6370 7360 8360 93601480 2480 3480   5380 6380 7380 8370 93701490 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

O2 = 1 O2= 2 O2= 3 O2= 4120 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)

SUMS FOR EACH COMBINATION OF F1O1UR AND O2NE

(219 * 280 = 61,320 SETS)

O2NE 

  120 123 124 125 126

F1O

1UR

1230 1230+120=13

501230+123=1353

1230+124=1354

1230+125=1355

1230+126=1356

1240 1240+120=13

601240+123=1363

1240+124=1364

1240+125=1365

1240+126=1366

1250 1250+120=13

701250+123=1373

1250+124=1374

1250+125=1375

1250+126=1376

1260 1260+120=13

801260+123=1383

1260+124=1384

1260+125=1385

1260+126=1386

1270 1270+120=13

901270+123=1393

1270+124=1394

1270+125=1395

1270+126=1396

1280 1280+120=14

001280+123=1403

1280+124=1404

1280+125=1405

1280+126=1406

1290 1290+120=14

101290+123=1413

1290+124=1414

1290+125=1415

1290+126=1416

1320 1320+120=14

401320+123=1443

1320+124=1444

1320+125=1445

1320+126=1446

1340 1340+120=14

601340+123=1463

1340+124=1464

1340+125=1465

1340+126=1466

1350 1350+120=14

701350+123=1473

1350+124=1474

1350+125=1475

1350+126=1476

1360 1360+120=14

801360+123=1483

1360+124=1484

1360+125=1485

1360+126=1486

1370 1370+120=14

901370+123=1493

1370+124=1494

1370+125=1495

1370+126=1496

1380 1380+120=15

001380+123=1503

1380+124=1504

1380+125=1505

1380+126=1506

1390 1390+120=15

101390+123=1513

1390+124=1514

1390+125=1515

1390+126=1516

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, VWhether O1 is distinct from N, E, I, VWhether U is distinct from N, E, I, VWhether R is distinct from N, E, I, V Whether N is distinct from I, VWhether E is distinct from I, VWhether I is distinct from V

If and only if all of the above are true, then the solution is valid for “FOUR + ONE = FIVE.”

LOGICAL STATEMENT CHECKS – SCREEN SHOT

CONCLUDING SPREADSHEET SNAPSHOT

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

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

METHOD FOR GENERATING SOLUTIONS – DISTINCT DIGITSFor 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

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

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

MORE WORD MATH PROBLEMS!

HERE + SHE = COMES

 

DAYS + TOO = SHORT

 

SATURN + URANUS = PLANETS

 

HEAD + HAND = KNEE

 

ORANGE + GREEN + RED = YELLOW