六年级奥数：第三讲 数论的方法技巧之一
1abb0qr
a=bq+r0≤rb
qr
r=0a=bqabb|abaab
2a|cb|cabab|c
31n
p1p2pka1a2ak1n
4n1
dn=a1+1a2+1ak+1
5nn+1xyx≤y-1
1n=an10n+an-110n-1+…+a0
2a=bq+r
42n=2mtt
1 11411019983
a3a2a1a0
1000a3+100a2+10a1+a0
10
10a3+a2+a1+a0=10a3+10a2+10a1+10a0
10
990a3+90a2-9a0=1998
110a3+10a2-a0=222
a0=8a2=1a3=2
218
a+b+c14
3 123100018
abcd4
a+b+c=18ma+b+d=18n
mn
c-d=18m-n
21818r
a=18a1+rb=18b1+rc=18c1+r
a1b1c1
a+b+c=18a1+b1+c1+3r
18|a+b+c18|3r6|rr=06121000=55×18+101210006244299656318
4 N5491N10
N
N572=49a3≥1a4≥2ak
a1+1a2+1an+1=10
a3+1≥2a4+1≥310=2×5
a1+1=a2+1=a5+1=…=an+1=1
a1=a2=a5=…an=0N257a4+1≥32
a3+1a4+1=10
a3+1=5a4+1=2a3+1=2a4+1=5N=52-1×75-1=5×74=12005
5 N123199819992000N21
210=1024211=2048200020002101024=210N102
5
6 11
900
xyz1110
x2+y2+z2≤10
1≤x≤30≤y≤30≤z≤3
100101102103110111112
120121122130200201202
211212220221300301310
100101
7 N235352NN2000
N
N|100N=10202550
N|1000N=100125200250500
4NN=1000125020002500500010001250
14
1k10k
2
8 3103331315233
13+15+23=5151=3×17
171317 3 31715
1610 179 188
2510 269 278
3410 359 368
377 (11)449 (12)458
(13)467 (14)557 (15)566
3+5+5=133+3+9=155+9+9=23
3359
9 12
100790919293949596
111312712
114115116117118119120
121122123124125126
12122334121312
m+2m+3m+1312m231312
2m为2341312m+2m+3m+4m+132341312
:
13+213+313+13
n=1×2×3×…×n12
m+1+2m+1+3m+1+m+1m
10 1005
1
2
3
4
5
1999
1711131131737
211133173
3111731
41173
51731
67311
73117
81173
41999=4×499+3199973117
11 100
N
1N=2aa=01232a
2N=2a+mm2a2m
N=100100=26+362×36=7272
12313111
12 12340
1111
223
1
2
3133-1=2
33
44133+1
55
9-3+1=5
9131+3+9=13
14
14+13=27
1+3+9+27=40
133233
13
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