Số 6303268125 có bao nhiêu ước số nguyên?
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Lời giải:
Báo saiÁp dụng công thức: Nếu số N được phân tích thành thừa số các số nguyên tố dạng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtaiabg2 % da9iaadchadaqhaaWcbaGaaGymaaqaaiaadUgadaWgaaadbaGaaGym % aaqabaaaaOGaaiOlaiaadchadaqhaaWcbaGaaGOmaaqaaiaadUgada % WgaaadbaGaaGOmaaqabaaaaOGaaiOlaiaac6cacaGGUaGaamiCamaa % DaaaleaacaWGUbaabaGaam4AamaaBaaameaacaWGUbaabeaaaaaaaa!463A! N = p_1^{{k_1}}.p_2^{{k_2}}...p_n^{{k_n}}\) thì số các ước nguyên dương bằng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Aaiabg2 % da9maabmaabaGaam4AamaaBaaaleaacaaIXaaabeaakiabgUcaRiaa % igdaaiaawIcacaGLPaaadaqadaqaaiaadUgadaWgaaWcbaGaaGOmaa % qabaGccqGHRaWkcaaIXaaacaGLOaGaayzkaaGaaiOlaiaac6cacaGG % UaWaaeWaaeaacaWGRbWaaSbaaSqaaiaad6gaaeqaaOGaey4kaSIaaG % ymaaGaayjkaiaawMcaaaaa!494E! k = \left( {{k_1} + 1} \right)\left( {{k_2} + 1} \right)...\left( {{k_n} + 1} \right)\). Do đó số các ước nguyên của N là 2k.
Với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtaiabg2 % da9iaaiAdacaaIZaGaaGimaiaaiodacaaIYaGaaGOnaiaaiIdacaaI % XaGaaGOmaiaaiwdacqGH9aqpcaaIZaWaaWbaaSqabeaacaaI1aaaaO % GaaiOlaiaaiwdadaahaaWcbeqaaiaaisdaaaGccaGGUaGaaG4namaa % CaaaleqabaGaaG4maaaakiaac6cacaaIXaGaaGymamaaCaaaleqaba % GaaGOmaaaaaaa!49CC! N = 6303268125 = {3^5}{.5^4}{.7^3}{.11^2}\) thì có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaac6 % cadaqadaqaaiaaiwdacqGHRaWkcaaIXaaacaGLOaGaayzkaaWaaeWa % aeaacaaI0aGaey4kaSIaaGymaaGaayjkaiaawMcaamaabmaabaGaaG % 4maiabgUcaRiaaigdaaiaawIcacaGLPaaadaqadaqaaiaaikdacqGH % RaWkcaaIXaaacaGLOaGaayzkaaGaeyypa0JaaG4naiaaikdacaaIWa % aaaa!4A2D! 2.\left( {5 + 1} \right)\left( {4 + 1} \right)\left( {3 + 1} \right)\left( {2 + 1} \right) = 720\) ước số nguyên.
