i.e.p. alfred nobel
TRANSCRIPT
ARITMÉTICA.CURSO:
TEMA:
DOCENTE:
I.E.P. ALFRED NOBELI.E.P. ALFRED NOBEL
MCD – MCM II
CHARLI JHON PAUCAR GAMARRA.
PRACTICA
1
MCD – MCM II
𝑄 3 2 4 2
𝑅 0
𝑑2𝑑
𝑑
9𝑑
2𝑑
20𝑑
9𝑑
69𝑑
69𝑑 − 20𝑑 = 245
49𝑑 = 245
𝑑 = 5
𝑆 = 69𝑑 + 20𝑑
𝑆 = 89𝑑
𝑆 = 445
2
MCD – MCM II
MCM(A;B) = m
MCD(A;B) = d
d.a = A
d.b = B
m = d.a.b
PESI :a y b
𝐴 ∗ 𝐵 = 960
𝑎. 𝑑 ∗ 𝑑. 𝑏 = 960
𝑚.𝑑 = 960
120 ∗ 𝑑 = 960
𝑑 = 8
MCD(A;B) = d = 8
3
MCD – MCM II
MCM(A;B) = 840
MCD(A;B) = 40
40a = A
40b = B
m = d.a.b
PESI :a y b
840 = 40ab
21 = ab
a= 7; b= 3
A+ B = 40(a+b)
A+ B = 40(10)
A+ B = 400
4
MCD – MCM II
𝑀𝐶𝐷 𝐴;8; 𝐶;𝐷 = 𝑀𝐶𝐷(𝑀𝐶𝐷 𝐴;8 ;𝑀𝐶𝐷 𝐶;𝐷 )
54 = 𝑀𝐶𝐷(15𝑘; 12𝑘)
54 = 3𝑘
18 = 𝑘
5
MCD – MCM II
𝑁 = 30𝑘
𝑀𝐶𝐷 10𝑘; 12𝑘; 35𝑘 = 24
𝑘 = 24
𝑁 = 30𝑘 = 30 ∗ 24 = 720
6
MCD – MCM II
𝑀𝐶𝑀 𝐴;𝐵 = 420
𝐴 𝑦 𝐵 𝑠𝑜𝑛 𝑝𝑒𝑠𝑖.
𝐴 ∗ 𝐵 = 420 = 22 ∗ 3 ∗ 5 ∗ 7
𝐴 ∗ 𝐵 = 12 ∗ 35
𝐴 + 𝐵 = 47
𝐴 = 12;𝐵 = 35
𝐵 − 𝐴 = 35 − 12 = 23
7
MCD – MCM II
𝑀𝐶𝐷 𝐴;𝐵;𝐶 = 6𝑀𝐶𝐷(8;12;20) − 1
𝑀𝐶𝐷 𝐴;𝐵; 𝐶 = 64 − 1
𝑀𝐶𝐷 𝐴;𝐵;𝐶 = 1296 − 1 = 1295
𝑆𝑐 = 1 + 2 + 9+ 5 = 17
8
MCD – MCM II
𝑄 2 3 4 2
𝑅 0
𝑑2𝑑
𝑑
9𝑑
2𝑑
29𝑑
9𝑑
67𝑑
67𝑑 + 29𝑑 = 288
96𝑑 = 288
𝑑 = 3
𝑀𝐶𝑀 67 ∗ 3; 29 ∗ 3 = 67 ∗ 29 ∗ 3
𝑀𝐶𝑀 67 ∗ 3; 29 ∗ 3 = 5829
𝑆𝑐 = 5 + 8 + 2 + 9 = 24
9
MCD – MCM II
MCM(A;B) = m
MCD(A;B) = d
d.a = A
d.b = B
m = d.a.b
PESI :a y b
𝑚
𝑑= 63
𝑎𝑏𝑑
𝑑= 63
𝑎𝑏 = 63
𝑎 = 7; 𝑏 = 9
𝑑 +𝑚 = 960
𝑑 + 𝑎𝑏𝑑 = 960
𝑑 1 + 𝑎𝑏 = 960
𝑑 64 = 960
𝑑 = 15
𝐴 + 𝐵 = 𝑎𝑑 + 𝑏𝑑
𝐴 + 𝐵 = 𝑑(𝑎 + 𝑏)
𝐴 + 𝐵 = 240
10
MCD – MCM II
MCM(A;B) = m
MCD(A;B) = d
d.a = A
d.b = B
m = d.a.b
PESI :a y b
𝐴 + 𝐵 = 325
𝑑(𝑎 + 𝑏) = 325
𝑎𝑑𝑏 = 1000
𝑎 + 𝑏
𝑎𝑏=13
40
𝑎 = 8; 𝑏 = 5
𝑑(𝑎 + 𝑏) = 325
𝑑(13) = 325
𝑑 = 25
𝐴 − 𝐵 = 8𝑑 − 5𝑑
𝐴 − 𝐵 = 3𝑑
𝐴 − 𝐵 = 3 ∗ 25
𝐴 − 𝐵 = 75
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